Chapter 9 Sequences And Series

(i) $a_n = 2n + 5$

To find the first three terms, we substitute $n=1, 2, 3$ into the given formula.

First term ($n=1$): $a_1 = 2(1) + 5 = 2 + 5 = 7$.

Second term ($n=2$): $a_2 = 2(2) + 5 = 4 + 5 = 9$.

Third term ($n=3$): $a_3 = 2(3) + 5 = 6 + 5 = 11$.

Therefore, the first three terms are 7, 9, 11.

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(ii) $a_n = \frac{n-3}{4}$

To find the first three terms, we substitute $n=1, 2, 3$ into the given formula.

First term ($n=1$): $a_1 = \frac{1-3}{4} = \frac{-2}{4} = -\frac{1}{2}$.

Second term ($n=2$): $a_2 = \frac{2-3}{4} = -\frac{1}{4}$.

Third term ($n=3$): $a_3 = \frac{3-3}{4} = \frac{0}{4} = 0$.

Therefore, the first three terms are $-\frac{1}{2}, -\frac{1}{4}, 0$.

Given:

The sequence is defined by the formula $a_n = (n-1)(2-n)(3+n)$.

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To Find:

The 20th term of the sequence ($a_{20}$).

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Solution:

To find the 20th term, we substitute $n=20$ into the given formula.

$a_{20} = (20 - 1)(2 - 20)(3 + 20)$

$a_{20} = (19)(-18)(23)$

$a_{20} = -342 \times 23$

$a_{20} = -7866$

Therefore, the 20th term of the sequence is -7866.

Given:

A sequence is defined recursively by:

$a_1 = 1$

$a_n = a_{n-1} + 2$ for $n \ge 2$.

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To Find:

The first few terms of the sequence.

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Solution:

This is a recursive formula, where each term is found by adding 2 to the previous term. This describes an arithmetic progression.

The first term is given as $a_1 = 1$.

The second term ($n=2$): $a_2 = a_{2-1} + 2 = a_1 + 2 = 1 + 2 = 3$.

The third term ($n=3$): $a_3 = a_{3-1} + 2 = a_2 + 2 = 3 + 2 = 5$.

The fourth term ($n=4$): $a_4 = a_{4-1} + 2 = a_3 + 2 = 5 + 2 = 7$.

The fifth term ($n=5$): $a_5 = a_{5-1} + 2 = a_4 + 2 = 7 + 2 = 9$.

The sequence is the set of odd positive integers.

The first five terms of the sequence are 1, 3, 5, 7, 9.

Given the n^th term of the sequence: $a_n = n (n + 2)$.

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To find the first five terms, we substitute $n=1, 2, 3, 4, 5$ into the formula for $a_n$.

For $n=1$: $a_1 = 1 (1 + 2) = 1 \times 3 = 3$.

For $n=2$: $a_2 = 2 (2 + 2) = 2 \times 4 = 8$.

For $n=3$: $a_3 = 3 (3 + 2) = 3 \times 5 = 15$.

For $n=4$: $a_4 = 4 (4 + 2) = 4 \times 6 = 24$.

For $n=5$: $a_5 = 5 (5 + 2) = 5 \times 7 = 35$.

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The first five terms of the sequence are 3, 8, 15, 24, and 35.

Given the n^th term of the sequence: $a_n = \frac{n}{n \;+\; 1}$.

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To find the first five terms, we substitute $n=1, 2, 3, 4, 5$ into the formula for $a_n$.

For $n=1$: $a_1 = \frac{1}{1 + 1} = \frac{1}{2}$.

For $n=2$: $a_2 = \frac{2}{2 + 1} = \frac{2}{3}$.

For $n=3$: $a_3 = \frac{3}{3 + 1} = \frac{3}{4}$.

For $n=4$: $a_4 = \frac{4}{4 + 1} = \frac{4}{5}$.

For $n=5$: $a_5 = \frac{5}{5 + 1} = \frac{5}{6}$.

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The first five terms of the sequence are $\mathbf{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \text{ and } \frac{5}{6}}$.

Given the n^th term of the sequence: $a_n = 2^n$.

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To find the first five terms, we substitute $n=1, 2, 3, 4, 5$ into the formula for $a_n$.

For $n=1$: $a_1 = 2^1 = 2$.

For $n=2$: $a_2 = 2^2 = 4$.

For $n=3$: $a_3 = 2^3 = 8$.

For $n=4$: $a_4 = 2^4 = 16$.

For $n=5$: $a_5 = 2^5 = 32$.

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The first five terms of the sequence are 2, 4, 8, 16, and 32.

Given the n^th term of the sequence: $a_n = \frac{2n \;-\; 3}{6}$.

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To find the first five terms, we substitute $n=1, 2, 3, 4, 5$ into the formula for $a_n$.

For $n=1$: $a_1 = \frac{2(1) - 3}{6} = \frac{2 - 3}{6} = \frac{-1}{6}$.

For $n=2$: $a_2 = \frac{2(2) - 3}{6} = \frac{4 - 3}{6} = \frac{1}{6}$.

For $n=3$: $a_3 = \frac{2(3) - 3}{6} = \frac{6 - 3}{6} = \frac{3}{6} = \frac{1}{2}$.

For $n=4$: $a_4 = \frac{2(4) - 3}{6} = \frac{8 - 3}{6} = \frac{5}{6}$.

For $n=5$: $a_5 = \frac{2(5) - 3}{6} = \frac{10 - 3}{6} = \frac{7}{6}$.

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The first five terms of the sequence are $\mathbf{-\frac{1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \text{ and } \frac{7}{6}}$.

Given the n^th term of the sequence: $a_n = (–1)^{n–1} 5^{n+1}$.

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To find the first five terms, we substitute $n=1, 2, 3, 4, 5$ into the formula for $a_n$.

For $n=1$: $a_1 = (-1)^{1-1} 5^{1+1} = (-1)^0 5^2 = 1 \times 25 = 25$.

For $n=2$: $a_2 = (-1)^{2-1} 5^{2+1} = (-1)^1 5^3 = -1 \times 125 = -125$.

For $n=3$: $a_3 = (-1)^{3-1} 5^{3+1} = (-1)^2 5^4 = 1 \times 625 = 625$.

For $n=4$: $a_4 = (-1)^{4-1} 5^{4+1} = (-1)^3 5^5 = -1 \times 3125 = -3125$.

For $n=5$: $a_5 = (-1)^{5-1} 5^{5+1} = (-1)^4 5^6 = 1 \times 15625 = 15625$.

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The first five terms of the sequence are 25, -125, 625, -3125, and 15625.

Given the n^th term of the sequence: $a_n = n\frac{n^{2}\;+\;5}{4}$.

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To find the first five terms, we substitute $n=1, 2, 3, 4, 5$ into the formula for $a_n$.

For $n=1$: $a_1 = 1 \times \frac{1^2 + 5}{4} = 1 \times \frac{1 + 5}{4} = 1 \times \frac{6}{4} = \frac{6}{4} = \frac{3}{2}$.

For $n=2$: $a_2 = 2 \times \frac{2^2 + 5}{4} = 2 \times \frac{4 + 5}{4} = 2 \times \frac{9}{4} = \frac{18}{4} = \frac{9}{2}$.

For $n=3$: $a_3 = 3 \times \frac{3^2 + 5}{4} = 3 \times \frac{9 + 5}{4} = 3 \times \frac{14}{4} = \frac{42}{4} = \frac{21}{2}$.

For $n=4$: $a_4 = 4 \times \frac{4^2 + 5}{4} = 4 \times \frac{16 + 5}{4} = 4 \times \frac{21}{4} = 21$.

For $n=5$: $a_5 = 5 \times \frac{5^2 + 5}{4} = 5 \times \frac{25 + 5}{4} = 5 \times \frac{30}{4} = \frac{150}{4} = \frac{75}{2}$.

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The first five terms of the sequence are $\mathbf{\frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21, \text{ and } \frac{75}{2}}$.

Given the n^th term of the sequence: $a_n = 4n - 3$.

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To find the 17^th term ($a_{17}$), we substitute $n=17$ into the formula:

$a_{17} = 4(17) - 3$

$a_{17} = 68 - 3$

$a_{17} = 65$

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To find the 24^th term ($a_{24}$), we substitute $n=24$ into the formula:

$a_{24} = 4(24) - 3$

$a_{24} = 96 - 3$

$a_{24} = 93$

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The 17^th term is $\mathbf{65}$ and the 24^th term is $\mathbf{93}$.

Given the n^th term of the sequence: $a_n = \frac{n^{2}}{2^{n}}$.

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To find the 7^th term ($a_{7}$), we substitute $n=7$ into the formula:

$a_{7} = \frac{7^2}{2^7}$

$a_{7} = \frac{49}{128}$

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The 7^th term is $\mathbf{\frac{49}{128}}$.

Given the n^th term of the sequence: $a_n = (–1)^{n – 1} n^3$.

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To find the 9^th term ($a_{9}$), we substitute $n=9$ into the formula:

$a_{9} = (-1)^{9-1} 9^3$

$a_{9} = (-1)^8 9^3$

$a_{9} = 1 \times (9 \times 9 \times 9)$

$a_{9} = 1 \times (81 \times 9)$

$a_{9} = 729$

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The 9^th term is $\mathbf{729}$.

Given the n^th term of the sequence: $a_n = \frac{n (n \;-\; 2)}{n \;+\; 3}$.

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To find the 20^th term ($a_{20}$), we substitute $n=20$ into the formula:

$a_{20} = \frac{20 (20 - 2)}{20 + 3}$

$a_{20} = \frac{20 (18)}{23}$

$a_{20} = \frac{360}{23}$

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The 20^th term is $\mathbf{\frac{360}{23}}$.

Given:

A sequence defined by $a_1 = 3$ and $a_n = 3a_{n-1} + 2$ for $n>1$.

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First Five Terms:

We are given the first term, $a_1 = 3$.

For $n=2: a_2 = 3a_1 + 2 = 3(3) + 2 = 9 + 2 = 11$.

For $n=3: a_3 = 3a_2 + 2 = 3(11) + 2 = 33 + 2 = 35$.

For $n=4: a_4 = 3a_3 + 2 = 3(35) + 2 = 105 + 2 = 107$.

For $n=5: a_5 = 3a_4 + 2 = 3(107) + 2 = 321 + 2 = 323$.

The first five terms of the sequence are 3, 11, 35, 107, 323.

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Corresponding Series:

The corresponding series is the sum of these terms.

The series is $3 + 11 + 35 + 107 + 323 + \dots$

Given:

A sequence defined by $a_1 = -1$ and $a_n = \frac{a_{n-1}}{n}$ for $n \ge 2$.

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First Five Terms:

We are given the first term, $a_1 = -1$.

For $n=2: a_2 = \frac{a_1}{2} = \frac{-1}{2}$.

For $n=3: a_3 = \frac{a_2}{3} = \frac{-1/2}{3} = -\frac{1}{6}$.

For $n=4: a_4 = \frac{a_3}{4} = \frac{-1/6}{4} = -\frac{1}{24}$.

For $n=5: a_5 = \frac{a_4}{5} = \frac{-1/24}{5} = -\frac{1}{120}$.

(Note: The terms are $a_n = -\frac{1}{n!}$)

The first five terms of the sequence are -1, -1/2, -1/6, -1/24, -1/120.

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Corresponding Series:

The corresponding series is the sum of these terms.

The series is $-1 - \frac{1}{2} - \frac{1}{6} - \frac{1}{24} - \frac{1}{120} - \dots$

Given:

A sequence defined by $a_1=2, a_2=2$, and $a_n = a_{n-1} - 1$ for $n>2$.

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First Five Terms:

We are given the first two terms: $a_1 = 2, a_2 = 2$.

For $n=3: a_3 = a_2 - 1 = 2 - 1 = 1$.

For $n=4: a_4 = a_3 - 1 = 1 - 1 = 0$.

For $n=5: a_5 = a_4 - 1 = 0 - 1 = -1$.

The first five terms of the sequence are 2, 2, 1, 0, -1.

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Corresponding Series:

The corresponding series is the sum of these terms.

The series is $2 + 2 + 1 + 0 - 1 + \dots$

Given:

The Fibonacci sequence defined by $a_1 = 1, a_2 = 1$, and $a_n = a_{n-1} + a_{n-2}$ for $n>2$.

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To Find:

The value of the ratio $\frac{a_{n+1}}{a_n}$ for $n = 1, 2, 3, 4, 5$.

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Solution:

First, we need to find the first six terms of the Fibonacci sequence to calculate the required ratios.

$a_1 = 1$

$a_2 = 1$

$a_3 = a_2 + a_1 = 1 + 1 = 2$

$a_4 = a_3 + a_2 = 2 + 1 = 3$

$a_5 = a_4 + a_3 = 3 + 2 = 5$

$a_6 = a_5 + a_4 = 5 + 3 = 8$

Now, we calculate the ratios:

For $n=1: \frac{a_{1+1}}{a_1} = \frac{a_2}{a_1} = \frac{1}{1} = 1$.

For $n=2: \frac{a_{2+1}}{a_2} = \frac{a_3}{a_2} = \frac{2}{1} = 2$.

For $n=3: \frac{a_{3+1}}{a_3} = \frac{a_4}{a_3} = \frac{3}{2}$.

For $n=4: \frac{a_{4+1}}{a_4} = \frac{a_5}{a_4} = \frac{5}{3}$.

For $n=5: \frac{a_{5+1}}{a_5} = \frac{a_6}{a_5} = \frac{8}{5}$.

The required values are $1, 2, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}$.

Given:

For an Arithmetic Progression (A.P.):

The $m^{\text{th}}$ term, $a_m = n$.

The $n^{\text{th}}$ term, $a_n = m$.

Condition: $m \neq n$.

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To Find:

The $p^{\text{th}}$ term, $a_p$.

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Solution:

Let the first term of the A.P. be 'a' and the common difference be 'd'.

The formula for the $k^{\text{th}}$ term of an A.P. is $a_k = a + (k-1)d$.

From the given information, we can write two equations:

Subtracting equation (ii) from equation (i):

$(a + (m-1)d) - (a + (n-1)d) = n - m$

$(m-1-n+1)d = n - m$

$(m-n)d = -(m-n)$

Since $m \neq n$, we can divide by $(m-n)$.

$d = -1$.

Now, substitute $d=-1$ into equation (i) to find $a$:

$a + (m-1)(-1) = n \implies a - m + 1 = n \implies a = m+n-1$.

Now we can find the $p^{\text{th}}$ term, $a_p$:

$a_p = a + (p-1)d$

$a_p = (m+n-1) + (p-1)(-1)$

$a_p = m+n-1 - p+1$

$a_p = m+n-p$.

The $p^{\text{th}}$ term is $m+n-p$.

Given:

The sum of $n$ terms of an A.P. is $S_n = nP + \frac{1}{2}n(n-1)Q$.

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To Find:

The common difference ($d$) of the A.P.

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Solution:

The common difference can be found by calculating $d = a_2 - a_1$. We can find the terms from the sum using the relation $a_n = S_n - S_{n-1}$.

First term ($a_1$):

$a_1 = S_1 = (1)P + \frac{1}{2}(1)(1-1)Q = P + 0 = P$.

Sum of first two terms ($S_2$):

$S_2 = (2)P + \frac{1}{2}(2)(2-1)Q = 2P + Q$.

Second term ($a_2$):

$a_2 = S_2 - S_1 = (2P+Q) - P = P+Q$.

Common difference ($d$):

$d = a_2 - a_1 = (P+Q) - P = Q$.

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Alternate Method:

The standard formula for the sum of an A.P. is $S_n = \frac{n}{2}[2a + (n-1)d]$, where 'a' is the first term and 'd' is the common difference.

$S_n = na + \frac{1}{2}n(n-1)d$.

Comparing this with the given formula $S_n = nP + \frac{1}{2}n(n-1)Q$, we can see that:

The first term $a$ corresponds to $P$.

The common difference $d$ corresponds to $Q$.

Therefore, the common difference is Q.

Given:

Let the two A.P.s have first terms $a_1, a_2$ and common differences $d_1, d_2$.

The ratio of the sum of their first $n$ terms is given by:

$\frac{S_{n,1}}{S_{n,2}} = \frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{3n+8}{7n+15}$.

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To Find:

The ratio of their 12th terms, $\frac{a_{12,1}}{a_{12,2}}$.

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Solution:

The 12th term of an A.P. is given by $a_{12} = a + (12-1)d = a + 11d$.

We need to find the ratio $\frac{a_1 + 11d_1}{a_2 + 11d_2}$.

We can rewrite the expression for the ratio of sums by dividing the numerator and denominator by 2:

$\frac{a_1 + \frac{n-1}{2}d_1}{a_2 + \frac{n-1}{2}d_2} = \frac{3n+8}{7n+15}$.

To find the ratio of the 12th terms, we need the coefficient of $d$ to be 11. So we set:

$\frac{n-1}{2} = 11 \implies n-1 = 22 \implies n=23$.

Now, we substitute $n=23$ into the given ratio of sums:

$\frac{a_1 + 11d_1}{a_2 + 11d_2} = \frac{3(23)+8}{7(23)+15} = \frac{69+8}{161+15} = \frac{77}{176}$.

Simplifying the fraction by dividing by 11:

$\frac{77 \div 11}{176 \div 11} = \frac{7}{16}$.

Therefore, the ratio of their 12th terms is 7 : 16.

Given:

Income in the first year ($a$) = $\textsf{₹} 3,00,000$.

Annual increase (common difference, $d$) = $\textsf{₹} 10,000$.

Number of years ($n$) = 20.

The annual incomes form an Arithmetic Progression.

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To Find:

The total amount received in 20 years ($S_{20}$).

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Solution:

We use the formula for the sum of the first $n$ terms of an A.P.:

$S_n = \frac{n}{2}[2a + (n-1)d]$.

Substitute the given values:

$S_{20} = \frac{20}{2}[2(300000) + (20-1)(10000)]$.

$S_{20} = 10[600000 + 19(10000)]$.

$S_{20} = 10[600000 + 190000]$.

$S_{20} = 10[790000] = 7,900,000$.

Therefore, the total amount he received in 20 years is $\textsf{₹} 79,00,000$.

Given:

We need to insert 6 numbers between 3 and 24 to form an A.P.

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To Find:

The 6 numbers to be inserted.

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Solution:

Let the 6 numbers be $A_1, A_2, A_3, A_4, A_5, A_6$.

The resulting A.P. will be $3, A_1, A_2, A_3, A_4, A_5, A_6, 24$.

In this A.P., the first term is $a = 3$.

The total number of terms is $6+2=8$.

The 8th term is $a_8 = 24$.

Using the formula $a_n = a + (n-1)d$ for the 8th term:

$a_8 = a + (8-1)d \implies 24 = 3 + 7d$.

$21 = 7d \implies d = 3$.

The common difference is 3. Now we can find the inserted numbers:

$A_1 = a + d = 3 + 3 = 6$.

$A_2 = a + 2d = 3 + 2(3) = 9$.

$A_3 = a + 3d = 3 + 3(3) = 12$.

$A_4 = a + 4d = 3 + 4(3) = 15$.

$A_5 = a + 5d = 3 + 5(3) = 18$.

$A_6 = a + 6d = 3 + 6(3) = 21$.

The 6 numbers are 6, 9, 12, 15, 18, 21.

To Find:

The sum of the odd integers from 1 to 2001.

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Solution:

The sequence of odd integers is 1, 3, 5, ..., 2001. This is an Arithmetic Progression (A.P.).

First term, $a = 1$.

Common difference, $d = 3 - 1 = 2$.

Last term, $a_n = 2001$.

First, we find the number of terms, $n$, using the formula $a_n = a + (n-1)d$.

$2001 = 1 + (n-1)2$

$2000 = (n-1)2$

$1000 = n-1 \implies n = 1001$.

Now, we find the sum using the formula $S_n = \frac{n}{2}(a + a_n)$.

$S_{1001} = \frac{1001}{2}(1 + 2001) = \frac{1001}{2}(2002) = 1001 \times 1001 = 1002001$.

The sum of the odd integers is 1,002,001.

To Find:

The sum of all natural numbers between 100 and 1000 that are multiples of 5.

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Solution:

The numbers are 105, 110, 115, ..., 995. This is an A.P.

First term, $a = 105$.

Common difference, $d = 5$.

Last term, $a_n = 995$.

First, find the number of terms, $n$.

$995 = 105 + (n-1)5 \implies 890 = (n-1)5 \implies 178 = n-1 \ $$ \implies n = 179$.

Now, find the sum using $S_n = \frac{n}{2}(a + a_n)$.

$S_{179} = \frac{179}{2}(105 + 995) = \frac{179}{2}(1100) = 179 \times 550 = 98,450$.

The sum of the numbers is 98,450.

Given:

In an A.P., the first term $a=2$.

Sum of first 5 terms ($S_5$) = $\frac{1}{4} \times$ (Sum of next 5 terms).

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To Show:

The 20th term, $a_{20} = -112$.

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Proof:

Let the common difference be $d$.

$S_5 = \frac{5}{2}[2(2) + (5-1)d] = \frac{5}{2}(4+4d) = 10+10d$.

The sum of the next 5 terms is $S_{10} - S_5$.

$S_{10} = \frac{10}{2}[2(2) + (10-1)d] = 5(4+9d) = 20+45d$.

Sum of next 5 terms = $(20+45d) - (10+10d) = 10+35d$.

From the given condition: $S_5 = \frac{1}{4}(S_{10} - S_5)$.

$10+10d = \frac{1}{4}(10+35d) \implies 40+40d = 10+35d \implies 5d = -30 \ $$ \implies d=-6$.

Now, find the 20th term:

$a_{20} = a + (20-1)d = 2 + 19(-6) = 2 - 114 = -112$.

Hence, proved.

Given:

A.P.: -6, -11/2, -5, ...

Sum of $n$ terms, $S_n = -25$.

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To Find:

The number of terms, $n$.

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Solution:

First term, $a = -6$.

Common difference, $d = (-\frac{11}{2}) - (-6) = -\frac{11}{2} + 6 = \frac{1}{2}$.

Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:

$-25 = \frac{n}{2}[2(-6) + (n-1)\frac{1}{2}] = \frac{n}{2}[-12 + \frac{n-1}{2}]$.

$-50 = n\left[\frac{-24+n-1}{2}\right] \implies -100 = n(n-25)$.

$n^2 - 25n + 100 = 0$.

Factoring the quadratic equation:

$(n-5)(n-20) = 0$.

The possible values for $n$ are $n=5$ or $n=20$. Both are valid.

The number of terms needed is 5 or 20.

Given:

In an A.P., $a_p = \frac{1}{q}$ and $a_q = \frac{1}{p}$.

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To Prove:

$S_{pq} = \frac{1}{2}(pq+1)$.

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Proof:

Let the first term be 'a' and common difference be 'd'.

$a_p = a + (p-1)d = \frac{1}{q}$. ...(i)

$a_q = a + (q-1)d = \frac{1}{p}$. ...(ii)

Subtracting (ii) from (i):

$(p-1-q+1)d = \frac{1}{q} - \frac{1}{p} = \frac{p-q}{pq}$.

$(p-q)d = \frac{p-q}{pq} \implies d = \frac{1}{pq}$.

Substitute $d$ in (i): $a + (p-1)\frac{1}{pq} = \frac{1}{q} \ $$ \implies a = \frac{1}{q} - \frac{p-1}{pq} = \frac{p-(p-1)}{pq} = \frac{1}{pq}$.

Now, find the sum of the first $pq$ terms:

$S_{pq} = \frac{pq}{2}[2a + (pq-1)d]$.

$S_{pq} = \frac{pq}{2}\left[2\left(\frac{1}{pq}\right) + (pq-1)\left(\frac{1}{pq}\right)\right]$.

$S_{pq} = \frac{pq}{2} \cdot \frac{1}{pq} [2 + (pq-1)] = \frac{1}{2}[2+pq-1] = \frac{1}{2}(pq+1)$.

Hence, proved.

Given:

A.P.: 25, 22, 19, ...

Sum of $n$ terms, $S_n = 116$.

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To Find:

The last term, $a_n$.

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Solution:

First term, $a = 25$.

Common difference, $d = 22 - 25 = -3$.

Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:

$116 = \frac{n}{2}[2(25) + (n-1)(-3)] = \frac{n}{2}[50 - 3n + 3] = \frac{n}{2}(53-3n)$.

$232 = 53n - 3n^2 \implies 3n^2 - 53n + 232 = 0$.

Using the quadratic formula, $n = \frac{53 \pm \sqrt{(-53)^2 - 4(3)(232)}}{2(3)} \ $$ = \frac{53 \pm \sqrt{2809 - 2784}}{6} \ $$ = \frac{53 \pm \sqrt{25}}{6} \ $$ = \frac{53 \pm 5}{6}$.

The possible values for $n$ are $n = \frac{58}{6}$ (not an integer) or $n = \frac{48}{6} = 8$.

So, there are 8 terms. We need to find the 8th term.

$a_8 = a + (8-1)d = 25 + 7(-3) = 25 - 21 = 4$.

The last term is 4.

Given:

The $k^{\text{th}}$ term of an A.P. is $a_k = 5k + 1$.

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To Find:

The sum of the first $n$ terms, $S_n$.

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Solution:

First, we find the first term and the common difference of the A.P.

First term ($a_1$): Substitute $k=1$ into the formula for $a_k$.

$a_1 = 5(1) + 1 = 6$.

Second term ($a_2$): Substitute $k=2$.

$a_2 = 5(2) + 1 = 11$.

Common difference ($d$): $d = a_2 - a_1 = 11 - 6 = 5$.

Now, we use the formula for the sum of the first $n$ terms of an A.P.:

$S_n = \frac{n}{2}[2a_1 + (n-1)d]$.

Substitute $a_1=6$ and $d=5$:

$S_n = \frac{n}{2}[2(6) + (n-1)5]$

$S_n = \frac{n}{2}[12 + 5n - 5]$

$S_n = \frac{n}{2}(5n + 7)$.

The sum to $n$ terms is $\frac{n(5n+7)}{2}$.

Given:

The sum of $n$ terms of an A.P. is $S_n = pn + qn^2$.

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To Find:

The common difference ($d$) of the A.P.

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Solution:

We can find the first and second terms of the A.P. to determine the common difference.

First term ($a_1$):

$a_1 = S_1 = p(1) + q(1)^2 = p + q$.

Sum of first two terms ($S_2$):

$S_2 = p(2) + q(2)^2 = 2p + 4q$.

Second term ($a_2$):

$a_2 = S_2 - S_1 = (2p+4q) - (p+q) = p + 3q$.

Common difference ($d$):

$d = a_2 - a_1 = (p+3q) - (p+q) = 2q$.

Therefore, the common difference is $2q$.

Given:

For two A.P.s, the ratio of the sum of their first $n$ terms is:

$\frac{S_{n,1}}{S_{n,2}} = \frac{5n+4}{9n+6}$.

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To Find:

The ratio of their 18th terms, $\frac{a_{18,1}}{a_{18,2}}$.

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Solution:

We know that $\frac{S_{n,1}}{S_{n,2}} = \frac{a_1 + \frac{n-1}{2}d_1}{a_2 + \frac{n-1}{2}d_2} = \frac{5n+4}{9n+6}$.

The ratio of the 18th terms is $\frac{a_{18,1}}{a_{18,2}} = \frac{a_1 + 17d_1}{a_2 + 17d_2}$.

To find this ratio, we set the coefficient of $d$ in the sum formula equal to 17:

$\frac{n-1}{2} = 17 \implies n-1 = 34 \implies n=35$.

Now, substitute $n=35$ into the given ratio of sums:

$\frac{a_{18,1}}{a_{18,2}} = \frac{5(35)+4}{9(35)+6} = \frac{175+4}{315+6} = \frac{179}{321}$.

Therefore, the ratio of their 18th terms is 179 : 321.

Given:

For an A.P., the sum of the first $p$ terms is equal to the sum of the first $q$ terms ($S_p = S_q$), where $p \neq q$.

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To Find:

The sum of the first $(p+q)$ terms, $S_{p+q}$.

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Solution:

Let the first term be $a$ and the common difference be $d$.

$S_p = S_q \implies \frac{p}{2}[2a+(p-1)d] = \frac{q}{2}[2a+(q-1)d]$.

$p(2a+pd-d) = q(2a+qd-d)$.

$2ap + p^2d - pd = 2aq + q^2d - qd$.

$2a(p-q) + (p^2-q^2)d - (p-q)d = 0$.

$2a(p-q) + (p-q)(p+q)d - (p-q)d = 0$.

Since $p \neq q$, we can divide by $(p-q)$:

$2a + (p+q)d - d = 0 \implies 2a + (p+q-1)d = 0$.

Now, we find the sum of the first $(p+q)$ terms:

$S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d]$.

From our previous result, the term in the square brackets is 0.

$S_{p+q} = \frac{p+q}{2}[0] = 0$.

Therefore, the sum of the first $(p+q)$ terms is 0.

Given:

For an A.P., $S_p = a$, $S_q = b$, $S_r = c$.

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To Prove:

$\frac{a}{p}(q-r) + \frac{b}{q}(r-p) + \frac{c}{r}(p-q) = 0$.

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Proof:

Let the first term be $A$ and the common difference be $D$.

$S_p = a = \frac{p}{2}[2A + (p-1)D] \implies \frac{a}{p} = \frac{1}{2}[2A + (p-1)D]$.

Similarly, $\frac{b}{q} = \frac{1}{2}[2A + (q-1)D]$ and $\frac{c}{r} = \frac{1}{2}[2A + (r-1)D]$.

Substitute these into the LHS of the expression to be proved:

LHS = $\frac{1}{2}[2A+(p-1)D](q-r) + \frac{1}{2}[2A+(q-1)D](r-p) \ $$ + \frac{1}{2}[2A+(r-1)D](p-q)$.

Factor out $\frac{1}{2}$ and group terms with $2A$ and $D$:

LHS

= $\frac{1}{2} \left[ 2A(q-r+r-p+p-q) + D((p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)) \right]$.

The term with $2A$ becomes $2A(0)=0$.

The term with $D$ becomes $D(pq-pr-q+r + qr-pq-r+p \ $$ + pr-qr-p+q) = D(0)=0$.

LHS = $\frac{1}{2}[0+0] = 0$.

Hence, proved.

Given:

$\frac{S_m}{S_n} = \frac{m^2}{n^2}$.

---

To Show:

$\frac{a_m}{a_n} = \frac{2m-1}{2n-1}$.

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Proof:

$\frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]} = \frac{m^2}{n^2} \implies \frac{2a+(m-1)d}{2a+(n-1)d} = \frac{m}{n}$.

$n(2a+md-d) = m(2a+nd-d) \ $$ \implies 2an+mnd-nd = 2am+mnd-md$.

$2a(n-m) = d(n-m)$. Since $m \neq n$, we have $d=2a$.

Now, we find the ratio of the terms:

$\frac{a_m}{a_n} = \frac{a+(m-1)d}{a+(n-1)d}$.

Substitute $d=2a$:

$\frac{a+(m-1)(2a)}{a+(n-1)(2a)} = \frac{a(1+2m-2)}{a(1+2n-2)} = \frac{2m-1}{2n-1}$.

Hence, proved.

Given:

Sum of $n$ terms of an A.P., $S_n = 3n^2 + 5n$.

The $m^{\text{th}}$ term, $a_m = 164$.

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To Find:

The value of $m$.

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Solution:

We can find the formula for the $n^{\text{th}}$ term, $a_n$, using the relation $a_n = S_n - S_{n-1}$.

First term ($a_1$): $a_1 = S_1 = 3(1)^2 + 5(1) = 3 + 5 = 8$.

Sum of first two terms ($S_2$): $S_2 = 3(2)^2 + 5(2) = 3(4) + 10 = 12 + 10 = 22$.

Second term ($a_2$): $a_2 = S_2 - S_1 = 22 - 8 = 14$.

Common difference ($d$): $d = a_2 - a_1 = 14 - 8 = 6$.

Now we have the first term $a=8$ and common difference $d=6$.

The formula for the $m^{\text{th}}$ term is $a_m = a + (m-1)d$.

We are given $a_m = 164$.

$164 = 8 + (m-1)6$.

$164 - 8 = (m-1)6$.

$156 = 6(m-1)$.

$m-1 = \frac{156}{6} = 26$.

$m = 26 + 1 = 27$.

Therefore, the value of $m$ is 27.

Given:

We need to insert five numbers between 8 and 26 to form an A.P.

---

To Find:

The five numbers to be inserted.

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Solution:

Let the five numbers be $A_1, A_2, A_3, A_4, A_5$.

The resulting A.P. is $8, A_1, A_2, A_3, A_4, A_5, 26$.

The first term is $a = 8$.

There are a total of $5+2=7$ terms. The last term is the 7th term, $a_7 = 26$.

Using the formula $a_n = a + (n-1)d$ for the 7th term:

$a_7 = a + (7-1)d \implies 26 = 8 + 6d$.

$18 = 6d \implies d = 3$.

The common difference is 3. The five numbers are:

$A_1 = 8+3=11$.

$A_2 = 11+3=14$.

$A_3 = 14+3=17$.

$A_4 = 17+3=20$.

$A_5 = 20+3=23$.

The five numbers are 11, 14, 17, 20, 23.

Given:

The expression $\frac{a^n + b^n}{a^{n-1} + b^{n-1}}$ is the Arithmetic Mean (A.M.) between $a$ and $b$.

---

To Find:

The value of $n$.

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Solution:

The A.M. of $a$ and $b$ is $\frac{a+b}{2}$.

We are given:

$\frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \frac{a+b}{2}$.

Cross-multiply:

$2(a^n + b^n) = (a+b)(a^{n-1} + b^{n-1})$.

$2a^n + 2b^n = a^n + ab^{n-1} + ba^{n-1} + b^n$.

$a^n + b^n - ab^{n-1} - a^{n-1}b = 0$.

$a^{n-1}(a-b) - b^{n-1}(a-b) = 0$.

$(a-b)(a^{n-1} - b^{n-1}) = 0$.

Since it is generally assumed that $a \neq b$, we must have:

$a^{n-1} - b^{n-1} = 0 \implies a^{n-1} = b^{n-1}$.

$\left(\frac{a}{b}\right)^{n-1} = 1 = \left(\frac{a}{b}\right)^0$.

Equating the exponents:

$n-1=0 \implies n=1$.

Therefore, the value of $n$ is 1.

Given:

$m$ numbers are inserted between 1 and 31 to form an A.P.

The ratio of the 7th inserted number to the $(m-1)$-th inserted number is $5:9$.

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Solution:

The A.P. has first term $a=1$ and total terms $n=m+2$. The last term is $a_{m+2}=31$.

$a_{m+2} = a + (m+2-1)d \implies 31 = 1 + (m+1)d \implies d = \frac{30}{m+1}$.

The 7th inserted number is the 8th term of the A.P., $A_8 = a+7d = 1 + 7\frac{30}{m+1}$.

The $(m-1)$-th inserted number is the $m$-th term of the A.P., $A_m = a+(m-1)d = 1 + (m-1)\frac{30}{m+1}$.

The ratio is given as:

$\frac{1 + \frac{210}{m+1}}{1 + \frac{30(m-1)}{m+1}} = \frac{5}{9} \implies \frac{\frac{m+1+210}{m+1}}{\frac{m+1+30m-30}{m+1}} = \frac{m+211}{31m-29} = \frac{5}{9}$.

$9(m+211) = 5(31m-29) \implies 9m + 1899 = 155m - 145$.

$2044 = 146m \implies m = \frac{2044}{146} = 14$.

Therefore, the value of $m$ is 14.

Given:

The instalments form an Arithmetic Progression (A.P.).

First instalment (first term), $a = \textsf{₹} 100$.

Monthly increase (common difference), $d = \textsf{₹} 5$.

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To Find:

The amount of the 30th instalment ($a_{30}$).

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Solution:

We use the formula for the $n^{\text{th}}$ term of an A.P.: $a_n = a + (n-1)d$.

For the 30th instalment, $n=30$.

$a_{30} = 100 + (30-1)5$

$a_{30} = 100 + 29 \times 5$

$a_{30} = 100 + 145 = 245$.

The amount he will pay in the 30th instalment is $\textsf{₹} 245$.

Given:

The interior angles of a polygon form an A.P.

Smallest angle (first term), $a = 120^\circ$.

Common difference, $d = 5^\circ$.

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To Find:

The number of sides of the polygon, $n$.

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Solution:

The sum of the interior angles of an $n$-sided polygon is $(n-2) \times 180^\circ$.

The sum of the $n$ angles in the A.P. is $S_n = \frac{n}{2}[2a + (n-1)d]$.

Equating the two formulas for the sum:

$(n-2) \times 180 = \frac{n}{2}[2(120) + (n-1)5]$.

$360(n-2) = n[240 + 5n - 5] \implies 360n - 720 = n(235+5n)$.

$360n - 720 = 235n + 5n^2 \implies 5n^2 - 125n + 720 = 0$.

Divide by 5: $n^2 - 25n + 144 = 0$.

Factoring: $(n-9)(n-16)=0$.

Possible values for $n$ are 9 and 16.

However, the largest interior angle of a convex polygon must be less than $180^\circ$.

If $n=16$, the largest angle is $a_{16} = 120 + (16-1)5 = 120 + 75 = 195^\circ$, which is not possible for a convex polygon.

If $n=9$, the largest angle is $a_9 = 120 + (9-1)5 = 120 + 40 = 160^\circ$, which is valid.

The number of sides of the polygon is 9.

Given:

The Geometric Progression (G.P.) is 5, 25, 125, ...

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To Find:

The 10th term ($a_{10}$) and the $n^{\text{th}}$ term ($a_n$).

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Solution:

First, we identify the first term and the common ratio of the G.P.

First term, $a = 5$.

Common ratio, $r = \frac{25}{5} = 5$.

The formula for the $n^{\text{th}}$ term of a G.P. is $a_n = ar^{n-1}$.

$n^{\text{th}}$ term:

$a_n = 5 \cdot (5)^{n-1} = 5^1 \cdot 5^{n-1} = 5^{1+n-1} = 5^n$.

10th term:

Using the formula for the $n^{\text{th}}$ term with $n=10$:

$a_{10} = 5^{10}$.

The 10th term is $5^{10}$ and the $n^{\text{th}}$ term is $5^n$.

Given:

The G.P. is 2, 8, 32, ...

A term in the G.P. is 131072.

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To Find:

The position ($n$) of the term 131072.

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Solution:

First term, $a = 2$.

Common ratio, $r = \frac{8}{2} = 4$.

Let the $n^{\text{th}}$ term be $a_n = 131072$.

Using the formula $a_n = ar^{n-1}$:

$131072 = 2 \cdot (4)^{n-1}$.

$65536 = 4^{n-1}$.

We need to express 65536 as a power of 4. We can write $4=2^2$.

$65536 = (2^2)^{n-1} = 2^{2n-2}$.

Also, $65536 = 2^{16}$.

Equating the powers of 2:

$2n-2 = 16 \implies 2n = 18 \implies n=9$.

Alternatively, $65536 = 4^8$.

$4^8 = 4^{n-1} \implies 8 = n-1 \implies n=9$.

The term 131072 is the 9th term of the G.P.

Given:

For a G.P., the 3rd term $a_3=24$ and the 6th term $a_6=192$.

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To Find:

The 10th term, $a_{10}$.

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Solution:

Let the first term be $a$ and the common ratio be $r$.

$a_3 = ar^2 = 24$. ...(i)

$a_6 = ar^5 = 192$. ...(ii)

Divide (ii) by (i): $\frac{ar^5}{ar^2} = \frac{192}{24} \implies r^3 = 8 \implies r=2$.

Substitute $r=2$ into (i): $a(2)^2 = 24 \implies 4a=24 \implies a=6$.

Now, find the 10th term:

$a_{10} = ar^{9} = 6 \cdot (2)^9 = 6 \cdot 512 = 3072$.

The 10th term is 3072.

Given:

The geometric series is $1 + \frac{2}{3} + \frac{4}{9} + \dots$.

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Solution:

First term, $a=1$.

Common ratio, $r = \frac{2/3}{1} = \frac{2}{3}$.

Sum of first n terms ($S_n$):

Using the formula $S_n = \frac{a(1-r^n)}{1-r}$ (since $|r|<1$):

$S_n = \frac{1(1 - (\frac{2}{3})^n)}{1 - \frac{2}{3}} = \frac{1 - (\frac{2}{3})^n}{1/3} = 3\left(1 - \left(\frac{2}{3}\right)^n\right)$.

The sum of the first n terms is $3\left(1 - \left(\frac{2}{3}\right)^n\right)$.

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Sum of first 5 terms ($S_5$):

Substitute $n=5$ into the formula for $S_n$:

$S_5 = 3\left(1 - \left(\frac{2}{3}\right)^5\right) = 3\left(1 - \frac{32}{243}\right) = 3\left(\frac{243-32}{243}\right) = 3\left(\frac{211}{243}\right) = \frac{211}{81}$.

The sum of the first 5 terms is $\frac{211}{81}$.

Given:

G.P.: 3, 3/2, 3/4, ...

Sum of $n$ terms, $S_n = \frac{3069}{512}$.

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Solution:

First term, $a=3$.

Common ratio, $r = \frac{3/2}{3} = \frac{1}{2}$.

Using the sum formula $S_n = \frac{a(1-r^n)}{1-r}$:

$\frac{3069}{512} = \frac{3(1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}} = \frac{3(1 - (\frac{1}{2})^n)}{1/2} = 6\left(1 - \left(\frac{1}{2}\right)^n\right)$.

$\frac{3069}{512 \times 6} = 1 - \left(\frac{1}{2}\right)^n \implies \frac{1023}{1024} = 1 - \left(\frac{1}{2}\right)^n$.

$\left(\frac{1}{2}\right)^n = 1 - \frac{1023}{1024} = \frac{1}{1024}$.

Since $1024 = 2^{10}$, we have $\left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^{10}$.

Therefore, $n=10$.

10 terms are needed.

Given:

For a G.P., the sum of the first three terms is $13/12$ and their product is $-1$.

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Solution:

Let the three terms of the G.P. be $\frac{a}{r}, a, ar$.

Product: $(\frac{a}{r})(a)(ar) = a^3 = -1 \implies a = -1$.

The terms are $\frac{-1}{r}, -1, -r$.

Sum: $\frac{-1}{r} - 1 - r = \frac{13}{12}$.

Multiply by $12r$: $-12 - 12r - 12r^2 = 13r$.

$12r^2 + 25r + 12 = 0$.

Factoring: $(4r+3)(3r+4) = 0$.

The possible values for the common ratio are $r = -\frac{3}{4}$ or $r = -\frac{4}{3}$.

If $r = -\frac{3}{4}$, the terms are $\frac{-1}{-3/4}, -1, -(-\frac{3}{4}) \implies \frac{4}{3}, -1, \frac{3}{4}$.

If $r = -\frac{4}{3}$, the terms are $\frac{-1}{-4/3}, -1, -(-\frac{4}{3}) \implies \frac{3}{4}, -1, \frac{4}{3}$.

The common ratio is -3/4 or -4/3, and the terms are (4/3, -1, 3/4) or (3/4, -1, 4/3).

To Find:

The sum of the sequence 7, 77, 777, ... to $n$ terms.

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Solution:

$S_n = 7 + 77 + 777 + \dots$ (n terms)

$S_n = 7(1 + 11 + 111 + \dots)$

Multiply and divide by 9:

$S_n = \frac{7}{9}(9 + 99 + 999 + \dots)$

$S_n = \frac{7}{9}((10-1) + (100-1) + (1000-1) + \dots + (10^n-1))$

$S_n = \frac{7}{9}((10+10^2+10^3+\dots+10^n) - (1+1+1+\dots+1))$

The first part is a G.P. with $a=10, r=10$. Its sum is $\frac{10(10^n-1)}{10-1} = \frac{10}{9}(10^n-1)$.

The second part is the sum of $n$ ones, which is $n$.

$S_n = \frac{7}{9}\left[\frac{10}{9}(10^n-1) - n\right]$.

The sum is $\frac{7}{9}\left[\frac{10}{9}(10^n-1) - n\right]$ or $\frac{70}{81}(10^n-1) - \frac{7n}{9}$.

Given:

The number of ancestors in preceding generations forms a sequence: 2, 4, 8, ...

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To Find:

The total number of ancestors in 10 generations.

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Solution:

The sequence is a Geometric Progression (G.P.).

First term, $a = 2$.

Common ratio, $r = 4/2 = 2$.

Number of terms (generations), $n = 10$.

We need to find the sum of the first 10 terms, $S_{10}$.

Using the formula $S_n = \frac{a(r^n-1)}{r-1}$ (since $r>1$):

$S_{10} = \frac{2(2^{10}-1)}{2-1} = 2(1024-1) = 2(1023) = 2046$.

The total number of ancestors is 2046.

Given:

We need to insert 3 numbers between 1 and 256 to form a G.P.

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Solution:

Let the three numbers be $g_1, g_2, g_3$. The G.P. is $1, g_1, g_2, g_3, 256$.

First term, $a=1$. Total terms, $n=5$. The 5th term, $a_5=256$.

Using the formula $a_n = ar^{n-1}$:

$a_5 = ar^4 \implies 256 = 1 \cdot r^4 \implies r^4 = 256$.

Since $4^4 = 256$ and $(-4)^4 = 256$, the possible common ratios are $r=4$ and $r=-4$.

Case 1: $r=4$

$g_1 = ar = 1(4)=4$, $g_2=ar^2=16$, $g_3=ar^3=64$. The numbers are 4, 16, 64.

Case 2: $r=-4$

$g_1 = ar = 1(-4)=-4$, $g_2=ar^2=16$, $g_3=ar^3=-64$. The numbers are -4, 16, -64.

The inserted numbers can be 4, 16, 64 or -4, 16, -64.

Given:

For two positive numbers $a$ and $b$:

Arithmetic Mean (A.M.) = 10.

Geometric Mean (G.M.) = 8.

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Solution:

A.M. = $\frac{a+b}{2} = 10 \implies a+b=20$. ...(i)

G.M. = $\sqrt{ab} = 8 \implies ab=64$. ...(ii)

We know the identity $(a-b)^2 = (a+b)^2 - 4ab$.

$(a-b)^2 = (20)^2 - 4(64) = 400 - 256 = 144$.

$a-b = \pm \sqrt{144} = \pm 12$. ...(iii)

Case 1: $a-b = 12$

Adding (i) and (iii): $(a+b)+(a-b)=20+12 \implies 2a=32 \implies a=16$.

Then $b = 20-a = 20-16 = 4$.

Case 2: $a-b = -12$

Adding (i) and (iii): $(a+b)+(a-b)=20-12 \implies 2a=8 \implies a=4$.

Then $b = 20-a = 20-4 = 16$.

In both cases, the numbers are 4 and 16.

The numbers are 4 and 16.

Given:

The geometric progression (G.P.) is $\frac{5}{2}$, $\frac{5}{4}$, $\frac{5}{8}$, ...

---

To Find:

The 20^th term and the n^th term of the given G.P.

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Solution:

The given sequence is $\frac{5}{2}$, $\frac{5}{4}$, $\frac{5}{8}$, ...

This is a G.P. with the first term $a = \frac{5}{2}$.

The common ratio, $r$, is found by dividing any term by its preceding term:

$r = \frac{\frac{5}{4}}{\frac{5}{2}} = \frac{5}{4} \times \frac{2}{5} = \frac{10}{20} = \frac{1}{2}$.

The common ratio is $r = \frac{1}{2}$.

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The formula for the n^th term of a G.P. with first term $a$ and common ratio $r$ is $a_n = ar^{n-1}$.

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To find the n^th term ($a_n$), substitute $a = \frac{5}{2}$ and $r = \frac{1}{2}$ into the formula:

$a_n = \frac{5}{2} \left(\frac{1}{2}\right)^{n-1}$

$a_n = \frac{5}{2^1} \times \frac{1^{n-1}}{2^{n-1}}$

$a_n = \frac{5}{2^1} \times \frac{1}{2^{n-1}}$

$a_n = \frac{5}{2^{1 + (n-1)}}$

$a_n = \frac{5}{2^n}$

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To find the 20^th term ($a_{20}$), substitute $n=20$ into the formula for $a_n$:

$a_{20} = \frac{5}{2^{20}}$

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Answer:

The n^th term of the G.P. is $a_n = \frac{5}{2^n}$.

The 20^th term of the G.P. is $a_{20} = \frac{5}{2^{20}}$.

Given:

The 8^th term of a G.P. is $a_8 = 192$.

The common ratio is $r = 2$.

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To Find:

The 12^th term of the G.P., $a_{12}$.

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Solution:

Let the first term of the G.P. be $a$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Using the given 8^th term ($n=8$) and common ratio ($r=2$):

$a_8 = ar^{8-1} = ar^7$

Substitute the value of $a_8$:

$192 = a(2)^7$

Calculate $2^7$:

$2^7 = 128$

So, $192 = a \times 128$

Solve for $a$:

$a = \frac{192}{128}$

Simplify the fraction:

$a = \frac{\cancel{192}^{3}}{\cancel{128}_{2}}$

$a = \frac{3}{2}$

The first term is $a = \frac{3}{2}$ and the common ratio is $r = 2$.

Now, find the 12^th term ($a_{12}$) using the formula $a_n = ar^{n-1}$ with $n=12$:

$a_{12} = ar^{12-1}$

$a_{12} = a r^{11}$

Substitute the values of $a$ and $r$:

$a_{12} = \frac{3}{2} \times 2^{11}$

$a_{12} = \frac{3}{2^1} \times 2^{11}$

$a_{12} = 3 \times 2^{11-1}$

$a_{12} = 3 \times 2^{10}$

Calculate $2^{10}$:

$2^{10} = 1024$

Substitute the value of $2^{10}$:

$a_{12} = 3 \times 1024$

$a_{12} = 3072$

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Answer:

The 12^th term of the G.P. is 3072.

Given:

The 5^th term of a G.P. is $a_5 = p$.

The 8^th term of the same G.P. is $a_8 = q$.

The 11^th term of the same G.P. is $a_{11} = s$.

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To Show:

$q^2 = ps$

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Proof:

Let the first term of the G.P. be $a$ and the common ratio be $r$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Using this formula, we can express the given terms:

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Consider the product $ps$ from Equations (1) and (3):

$ps = (ar^4)(ar^{10})$

$ps = a \cdot a \cdot r^4 \cdot r^{10}$

Using the exponent rule $x^m \cdot x^n = x^{m+n}$:

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Now consider $q^2$ from Equation (2):

$q = ar^7$

$q^2 = (ar^7)^2$

Using the exponent rule $(xy)^m = x^m y^m$ and $(x^m)^n = x^{mn}$:

---

Comparing Equation (4) and Equation (5), we see that $ps = a^2 r^{14}$ and $q^2 = a^2 r^{14}$.

Therefore, $q^2 = ps$.

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Answer:

Hence, it is shown that $q^2 = ps$.

Given:

The first term of the G.P. is $a_1 = -3$.

The 4^th term is the square of the second term, i.e., $a_4 = (a_2)^2$.

---

To Find:

The 7^th term of the G.P., $a_7$.

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Solution:

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$, where $a$ is the first term and $r$ is the common ratio.

We are given the first term $a_1 = a = -3$.

The second term is $a_2 = ar^{2-1} = ar = -3r$.

The fourth term is $a_4 = ar^{4-1} = ar^3 = -3r^3$.

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According to the given condition, the 4^th term is the square of the second term:

Substitute the expressions for $a_2$ and $a_4$ into Equation (1):

$-3r^3 = (-3r)^2$

$-3r^3 = (-3)^2 r^2$

$-3r^3 = 9r^2$

Move all terms to one side to form an equation equal to zero:

$9r^2 + 3r^3 = 0$

Factor out the common term $3r^2$:

$3r^2(3 + r) = 0$

This equation holds true if either $3r^2 = 0$ or $3 + r = 0$.

Case 1: $3r^2 = 0 \implies r^2 = 0 \implies r = 0$.

Case 2: $3 + r = 0 \implies r = -3$.

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For a standard geometric progression, the common ratio $r$ must be non-zero. If $r=0$ and $a \neq 0$, the sequence is $a, 0, 0, 0, \dots$. The ratio of terms after the first term becomes undefined ($0/0$). Therefore, we consider the valid common ratio $r = -3$.

---

The common ratio is $r = -3$. The first term is $a = -3$.

We need to find the 7^th term ($a_7$). Using the formula $a_n = ar^{n-1}$ with $n=7$:

$a_7 = ar^{7-1}$

$a_7 = ar^6$

Substitute the values of $a = -3$ and $r = -3$:

$a_7 = (-3)(-3)^6$

$a_7 = (-3)(729)$

Perform the multiplication:

$a_7 = -2187$

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Answer:

The 7^th term of the G.P. is -2187.

(a) For the G.P. 2, $2\sqrt{2}$, 4, … is 128?

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Given:

The geometric progression (G.P.) is 2, $2\sqrt{2}$, 4, ...

A term in the G.P. is 128.

---

To Find:

The term number of the term 128.

---

Solution:

The given sequence is 2, $2\sqrt{2}$, 4, ...

This is a G.P. with the first term $a = 2$.

The common ratio, $r$, is found by dividing any term by its preceding term:

$r = \frac{2\sqrt{2}}{2} = \sqrt{2}$.

Let the n^th term of the G.P. be $a_n$. We are given that $a_n = 128$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Substitute the known values into the formula:

$128 = 2 (\sqrt{2})^{n-1}$

Divide both sides by 2:

$\frac{128}{2} = (\sqrt{2})^{n-1}$

$64 = (\sqrt{2})^{n-1}$

Express both sides with the same base, 2. We know that $64 = 2^6$ and $\sqrt{2} = 2^{1/2}$.

$2^6 = (2^{1/2})^{n-1}$

Using the exponent rule $(x^m)^n = x^{mn}$:

$2^6 = 2^{\frac{n-1}{2}}$

Since the bases are equal, the exponents must be equal:

$6 = \frac{n-1}{2}$

Multiply both sides by 2:

$12 = n - 1$

Add 1 to both sides to solve for $n$:

$n = 12 + 1$

$n = 13$

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Answer:

The 13^th term of the G.P. is 128.

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(b) For the G.P. $\sqrt{3}$, 3, 3$\sqrt{3}$ , … is 729?

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Given:

The geometric progression (G.P.) is $\sqrt{3}$, 3, $3\sqrt{3}$, ...

A term in the G.P. is 729.

---

To Find:

The term number of the term 729.

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Solution:

The given sequence is $\sqrt{3}$, 3, $3\sqrt{3}$, ...

This is a G.P. with the first term $a = \sqrt{3}$.

The common ratio, $r$, is found by dividing any term by its preceding term:

$r = \frac{3}{\sqrt{3}} = \sqrt{3}$.

Let the n^th term of the G.P. be $a_n$. We are given that $a_n = 729$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Substitute the known values into the formula:

$729 = \sqrt{3} (\sqrt{3})^{n-1}$

Using the exponent rule $x^m \cdot x^n = x^{m+n}$ ($ \sqrt{3} = (\sqrt{3})^1$):

$729 = (\sqrt{3})^{1 + (n-1)}$

$729 = (\sqrt{3})^n$

Express both sides with the same base, 3. We know that $729 = 3^6$ and $\sqrt{3} = 3^{1/2}$.

$3^6 = (3^{1/2})^n$

Using the exponent rule $(x^m)^n = x^{mn}$:

$3^6 = 3^{\frac{n}{2}}$

Since the bases are equal, the exponents must be equal:

$6 = \frac{n}{2}$

Multiply both sides by 2:

$n = 12$

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Answer:

The 12^th term of the G.P. is 729.

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(c) For the G.P. $\frac{1}{3}$ , $\frac{1}{9}$ , $\frac{1}{27}$, … is $\frac{1}{19683}$?

---

Given:

The geometric progression (G.P.) is $\frac{1}{3}$, $\frac{1}{9}$, $\frac{1}{27}$, ...

A term in the G.P. is $\frac{1}{19683}$.

---

To Find:

The term number of the term $\frac{1}{19683}$.

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Solution:

The given sequence is $\frac{1}{3}$, $\frac{1}{9}$, $\frac{1}{27}$, ...

This is a G.P. with the first term $a = \frac{1}{3}$.

The common ratio, $r$, is found by dividing any term by its preceding term:

$r = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{9} \times 3 = \frac{3}{9} = \frac{1}{3}$.

Let the n^th term of the G.P. be $a_n$. We are given that $a_n = \frac{1}{19683}$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Substitute the known values into the formula:

$\frac{1}{19683} = \frac{1}{3} \left(\frac{1}{3}\right)^{n-1}$

Using the exponent rule $x^m \cdot x^n = x^{m+n}$:

$\frac{1}{19683} = \left(\frac{1}{3}\right)^{1 + (n-1)}$

$\frac{1}{19683} = \left(\frac{1}{3}\right)^n$

To find $n$, we need to determine what power of 3 gives 19683 in the denominator. This is equivalent to finding $n$ such that $3^n = 19683$.

Let's calculate powers of 3:

$3^1 = 3$

$3^2 = 9$

$3^3 = 27$

$3^4 = 81$

$3^5 = 243$

$3^6 = 729$

$3^7 = 2187$

$3^8 = 6561$

$3^9 = 19683$

So, $19683 = 3^9$.

The equation is $\frac{1}{3^9} = \left(\frac{1}{3}\right)^n$, which means $\left(\frac{1}{3}\right)^9 = \left(\frac{1}{3}\right)^n$.

Since the bases are equal, the exponents must be equal:

$n = 9$

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Answer:

The 9^th term of the G.P. is $\frac{1}{19683}$.

Given:

The three numbers $-\frac{2}{7}$, $x$, $-\frac{7}{2}$ are in Geometric Progression (G.P.).

---

To Find:

The values of $x$ for which the given numbers form a G.P.

---

Solution:

For three numbers, $a$, $b$, and $c$ to be in G.P., the ratio of consecutive terms must be the same. That is, $\frac{b}{a} = \frac{c}{b}$.

This property can be written as $b^2 = ac$. The middle term $b$ is the geometric mean of the other two terms $a$ and $c$.

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In this case, the three terms are $a = -\frac{2}{7}$, $b = x$, and $c = -\frac{7}{2}$.

Using the property $b^2 = ac$, we have:

$x^2 = \left(-\frac{2}{7}\right) \times \left(-\frac{7}{2}\right)$

Multiply the fractions on the right side:

$x^2 = \left(\frac{-2 \times -7}{7 \times 2}\right)$

$x^2 = \frac{14}{14}$

$x^2 = 1$

To find the values of $x$, take the square root of both sides:

$x = \pm \sqrt{1}$

$x = \pm 1$

So, there are two possible values for $x$: $1$ and $-1$.

---

We can verify these values:

If $x = 1$, the sequence is $-\frac{2}{7}$, 1, $-\frac{7}{2}$.

The ratio of the second term to the first term is $\frac{1}{-2/7} = 1 \times (-\frac{7}{2}) = -\frac{7}{2}$.

The ratio of the third term to the second term is $\frac{-7/2}{1} = -\frac{7}{2}$.

Since the ratios are equal ($-\frac{7}{2}$), the sequence is a G.P. with $r = -\frac{7}{2}$.

---

If $x = -1$, the sequence is $-\frac{2}{7}$, -1, $-\frac{7}{2}$.

The ratio of the second term to the first term is $\frac{-1}{-2/7} = -1 \times (-\frac{7}{2}) = \frac{7}{2}$.

The ratio of the third term to the second term is $\frac{-7/2}{-1} = \frac{7}{2}$.

Since the ratios are equal ($\frac{7}{2}$), the sequence is a G.P. with $r = \frac{7}{2}$.

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Answer:

The values of x for which the numbers $-\frac{2}{7}$, x, $-\frac{7}{2}$ are in G.P. are $x = 1$ and $x = -1$.

Given:

The geometric progression (G.P.) is 0.15, 0.015, 0.0015, ...

The number of terms is $n = 20$.

---

To Find:

The sum of the first 20 terms of the given G.P.

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Solution:

The given sequence is 0.15, 0.015, 0.0015, ...

The first term is $a = 0.15$.

The common ratio, $r$, is found by dividing a term by its preceding term:

$r = \frac{0.015}{0.15} = \frac{15/1000}{15/100} = \frac{15}{1000} \times \frac{100}{15} = \frac{1}{10} = 0.1$.

Since $|r| = |0.1| < 1$, the formula for the sum of the first $n$ terms of a G.P. is $S_n = \frac{a(1 - r^n)}{1 - r}$.

We need to find the sum of the first 20 terms, so $n=20$. Substitute the values of $a$, $r$, and $n$ into the formula:

$S_{20} = \frac{0.15(1 - (0.1)^{20})}{1 - 0.1}$

Simplify the denominator:

$1 - 0.1 = 0.9$

So the expression becomes:

$S_{20} = \frac{0.15(1 - (0.1)^{20})}{0.9}$

Simplify the fraction $\frac{0.15}{0.9}$:

$\frac{0.15}{0.9} = \frac{15/100}{9/10} = \frac{15}{100} \times \frac{10}{9} = \frac{15 \times 10}{100 \times 9} = \frac{150}{900} = \frac{15}{90} = \frac{1}{6}$.

The expression for $S_{20}$ is:

$S_{20} = \frac{1}{6} (1 - (0.1)^{20})$

This can also be written as:

$S_{20} = \frac{1}{6} (1 - 10^{-20})$

Or, distributing the $\frac{1}{6}$:

$S_{20} = \frac{1}{6} - \frac{1}{6} \times 10^{-20}$

---

Answer:

The sum of the first 20 terms of the G.P. is $\frac{1}{6} (1 - (0.1)^{20})$ or $\frac{1}{6}(1 - 10^{-20})$.

Given:

The geometric progression (G.P.) is $\sqrt{7}$, $\sqrt{21}$, $3\sqrt{7}$, ...

The number of terms is $n$.

---

To Find:

The sum of the first n terms of the given G.P.

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Solution:

The given sequence is $\sqrt{7}$, $\sqrt{21}$, $3\sqrt{7}$, ...

The first term is $a = \sqrt{7}$.

The common ratio, $r$, is found by dividing a term by its preceding term:

$r = \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{\frac{21}{7}} = \sqrt{3}$.

The common ratio is $r = \sqrt{3}$. Since $|r| = |\sqrt{3}| > 1$, the formula for the sum of the first $n$ terms of a G.P. is $S_n = \frac{a(r^n - 1)}{r - 1}$.

Substitute the values of $a$ and $r$ into the formula for $S_n$:

$S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)}{\sqrt{3} - 1}$

To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} + 1$:

$S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$

$S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2}$

$S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)(\sqrt{3} + 1)}{3 - 1}$

$S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)(\sqrt{3} + 1)}{2}$

We can also write $(\sqrt{3})^n$ as $(3^{1/2})^n = 3^{n/2}$.

$S_n = \frac{\sqrt{7}(3^{n/2} - 1)(\sqrt{3} + 1)}{2}$

---

Answer:

The sum of the first n terms of the G.P. is $S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)(\sqrt{3} + 1)}{2}$.

Given:

The geometric progression (G.P.) is 1, – a, a², – a³, ...

The number of terms is $n$.

Condition: $a \neq -1$.

---

To Find:

The sum of the first n terms of the given G.P.

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Solution:

The given sequence is 1, – a, a², – a³, ...

The first term is $A = 1$.

The common ratio, $r$, is found by dividing a term by its preceding term:

$r = \frac{-\text{a}}{1} = -\text{a}$.

The common ratio is $r = -a$.

The condition $a \neq -1$ implies $r = -a \neq -(-1) = 1$. Thus, $r \neq 1$.

The formula for the sum of the first n terms of a G.P. with first term $A$ and common ratio $r \neq 1$ is $S_n = \frac{A(1 - r^n)}{1 - r}$.

Substitute the values $A=1$ and $r=-a$ into the formula for $S_n$:

$S_n = \frac{1(1 - (-a)^n)}{1 - (-a)}$

$S_n = \frac{1 - (-a)^n}{1 + a}$

We can write $(-a)^n$ as $(-1 \times a)^n = (-1)^n a^n$.

So the sum is:

$S_n = \frac{1 - (-1)^n a^n}{1 + a}$

---

Answer:

The sum of the first n terms of the G.P. is $S_n = \frac{1 - (-1)^n a^n}{1 + a}$.

Given:

The geometric progression (G.P.) is x³, x⁵, x⁷, ...

The number of terms is $n$.

Condition: $x \neq \pm 1$.

---

To Find:

The sum of the first n terms of the given G.P.

---

Solution:

The given sequence is x³, x⁵, x⁷, ...

The first term is $a = x^3$.

The common ratio, $r$, is found by dividing a term by its preceding term:

$r = \frac{x^5}{x^3} = x^{5-3} = x^2$.

We can verify this with the next pair of terms: $r = \frac{x^7}{x^5} = x^{7-5} = x^2$.

The common ratio is $r = x^2$.

The condition $x \neq \pm 1$ implies $x^2 \neq (\pm 1)^2$, so $x^2 \neq 1$. Thus, the common ratio $r \neq 1$.

The formula for the sum of the first n terms of a G.P. with first term $a$ and common ratio $r \neq 1$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.

Substitute the values $a=x^3$ and $r=x^2$ into the formula for $S_n$:

$S_n = \frac{x^3((x^2)^n - 1)}{x^2 - 1}$

Using the exponent rule $(p^m)^n = p^{mn}$:

$(x^2)^n = x^{2 \times n} = x^{2n}$.

So the sum is:

$S_n = \frac{x^3(x^{2n} - 1)}{x^2 - 1}$

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Answer:

The sum of the first n terms of the G.P. is $S_n = \frac{x^3(x^{2n} - 1)}{x^2 - 1}$.

Given:

The summation expression $\sum\limits_{k=1}^{11} (2 + 3^k)$.

---

To Evaluate:

The value of the given summation.

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Solution:

The summation can be split into two parts using the property of summation $\sum\limits (a_k + b_k) = \sum\limits a_k + \sum\limits b_k$:

$\sum\limits_{k=1}^{11} (2 + 3^k) = \sum\limits_{k=1}^{11} 2 + \sum\limits_{k=1}^{11} 3^k$

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Consider the first part of the summation: $\sum\limits_{k=1}^{11} 2$.

This is the sum of a constant term (2) repeated 11 times:

$\sum\limits_{k=1}^{11} 2 = 2 + 2 + \dots + 2$ (11 times)

$\sum\limits_{k=1}^{11} 2 = 11 \times 2 = 22$

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Consider the second part of the summation: $\sum\limits_{k=1}^{11} 3^k$.

This expands to $3^1 + 3^2 + 3^3 + \dots + 3^{11}$.

This is a finite geometric series with:

First term, $a = 3^1 = 3$.

Common ratio, $r = \frac{3^2}{3^1} = 3$.

Number of terms, $n = 11$.

Since the common ratio $r = 3 \neq 1$, the formula for the sum of the first $n$ terms of a G.P. is $S_n = \frac{a(r^n - 1)}{r - 1}$.

Substitute the values $a=3$, $r=3$, and $n=11$ into the formula for the sum of this G.P.:

$\sum\limits_{k=1}^{11} 3^k = \frac{3(3^{11} - 1)}{3 - 1}$

$= \frac{3(3^{11} - 1)}{2}$

Calculate $3^{11}$:

$3^{11} = 177147$

Substitute the value of $3^{11}$:

$= \frac{3(177147 - 1)}{2}$

$= \frac{3(177146)}{2}$

$= 3 \times \frac{177146}{2}$

$= 3 \times 88573$

$= 265719$

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Now, add the sums of the two parts:

$\sum\limits_{k=1}^{11} (2 + 3^k) = \sum\limits_{k=1}^{11} 2 + \sum\limits_{k=1}^{11} 3^k$

$= 22 + 265719$

$= 265741$

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Answer:

The value of the summation $\sum\limits_{k=1}^{11} (2 + 3^k)$ is 265741.

Given:

The sum of the first three terms of a G.P. is $\frac{39}{10}$.

The product of the first three terms of the G.P. is 1.

---

To Find:

The common ratio and the first three terms of the G.P.

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Solution:

Let the first three terms of the G.P. be $\frac{a}{r}$, $a$, and $ar$, where $a$ is the first term and $r$ is the common ratio. (Note: Using this representation simplifies the product calculation. The terms in standard notation would be $a_1 = \frac{a}{r}$, $a_2 = a$, $a_3 = ar$).

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According to the given information, the product of the first three terms is 1:

$\left(\frac{a}{r}\right) \times (a) \times (ar) = 1$

$a^{1} \times a^{1} \times a^{1} \times r^{-1} \times r^{1} = 1$

$a^{1+1+1} \times r^{-1+1} = 1$

$a^3 \times r^0 = 1$

Since $r^0 = 1$ (assuming $r \neq 0$), we have:

Taking the cube root of both sides, we get $a = 1$.

So, the second term of the G.P. is 1. The three terms are $\frac{1}{r}$, $1$, and $r$.

---

According to the given information, the sum of the first three terms is $\frac{39}{10}$:

$\frac{1}{r} + 1 + r = \frac{39}{10}$

Combine the terms on the left side by finding a common denominator $r$:

$\frac{1}{r} + \frac{r}{r} + \frac{r^2}{r} = \frac{39}{10}$

$\frac{1 + r + r^2}{r} = \frac{39}{10}$

Cross-multiply:

$10(1 + r + r^2) = 39r$

Expand the left side:

$10 + 10r + 10r^2 = 39r$

Move all terms to one side to form a quadratic equation in $r$:

$10r^2 + 10r - 39r + 10 = 0$

---

We can solve this quadratic equation by factoring. We look for two numbers that multiply to $10 \times 10 = 100$ and add up to -29. These numbers are -4 and -25.

Rewrite the middle term:

$10r^2 - 4r - 25r + 10 = 0$

Group the terms and factor by grouping:

$(10r^2 - 4r) + (-25r + 10) = 0$

$2r(5r - 2) - 5(5r - 2) = 0$

Factor out the common binomial $(5r - 2)$:

$(5r - 2)(2r - 5) = 0$

This gives two possible values for the common ratio $r$:

Case 1: $5r - 2 = 0 \implies 5r = 2 \implies r = \frac{2}{5}$

Case 2: $2r - 5 = 0 \implies 2r = 5 \implies r = \frac{5}{2}$

---

Now, we find the three terms for each case, using $a = 1$ and the respective common ratio $r$. The terms are $\frac{a}{r}, a, ar$, which simplifies to $\frac{1}{r}, 1, r$ since $a=1$.

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Case 1: Common ratio $r = \frac{2}{5}$

The terms are:

First term: $\frac{1}{r} = \frac{1}{2/5} = 1 \times \frac{5}{2} = \frac{5}{2}$

Second term: $1$

Third term: $r = \frac{2}{5}$

The terms are $\frac{5}{2}, 1, \frac{2}{5}$.

Check sum: $\frac{5}{2} + 1 + \frac{2}{5} = \frac{25}{10} + \frac{10}{10} + \frac{4}{10} = \frac{25+10+4}{10} = \frac{39}{10}$. (Correct)

Check product: $\frac{5}{2} \times 1 \times \frac{2}{5} = 1$. (Correct)

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Case 2: Common ratio $r = \frac{5}{2}$

The terms are:

First term: $\frac{1}{r} = \frac{1}{5/2} = 1 \times \frac{2}{5} = \frac{2}{5}$

Second term: $1$

Third term: $r = \frac{5}{2}$

The terms are $\frac{2}{5}, 1, \frac{5}{2}$.

Check sum: $\frac{2}{5} + 1 + \frac{5}{2} = \frac{4}{10} + \frac{10}{10} + \frac{25}{10} = \frac{4+10+25}{10} = \frac{39}{10}$. (Correct)

Check product: $\frac{2}{5} \times 1 \times \frac{5}{2} = 1$. (Correct)

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Both cases satisfy the given conditions.

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Answer:

The common ratio is $\frac{2}{5}$ or $\frac{5}{2}$.

The terms of the G.P. are $\frac{5}{2}, 1, \frac{2}{5}$ (when $r = \frac{2}{5}$) or $\frac{2}{5}, 1, \frac{5}{2}$ (when $r = \frac{5}{2}$).

Given:

The geometric progression (G.P.) is 3, $3^2$, $3^3$, ... which is 3, 9, 27, ...

The sum of the first n terms is $S_n = 120$.

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To Find:

The number of terms, $n$, needed to give the sum 120.

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Solution:

The given sequence is 3, 9, 27, ...

This is a G.P. with the first term $a = 3$.

The common ratio, $r$, is found by dividing any term by its preceding term:

$r = \frac{9}{3} = 3$.

The common ratio is $r = 3$. Since $r \neq 1$, the formula for the sum of the first $n$ terms of a G.P. is $S_n = \frac{a(r^n - 1)}{r - 1}$.

We are given $S_n = 120$. Substitute the values of $a$, $r$, and $S_n$ into the formula:

$120 = \frac{3(3^n - 1)}{3 - 1}$

Simplify the denominator:

$3 - 1 = 2$

So the equation becomes:

$120 = \frac{3(3^n - 1)}{2}$

Multiply both sides by 2:

$120 \times 2 = 3(3^n - 1)$

$240 = 3(3^n - 1)$

Divide both sides by 3:

$\frac{240}{3} = 3^n - 1$

$80 = 3^n - 1$

Add 1 to both sides to isolate the term with $n$:

$80 + 1 = 3^n$

$81 = 3^n$

We need to express 81 as a power of 3:

$81 = 3 \times 3 \times 3 \times 3 = 3^4$.

So, we have:

$3^4 = 3^n$

Since the bases are equal, the exponents must be equal:

$n = 4$

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Answer:

4 terms of the G.P. are needed to give the sum 120.

Given:

The sum of the first three terms of a G.P. is $S_3 = 16$.

The sum of the next three terms (i.e., the 4^th, 5^th, and 6^th terms) is 128.

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To Find:

The first term ($a$), the common ratio ($r$), and the sum to n terms ($S_n$) of the G.P.

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Solution:

Let the first term of the G.P. be $a$ and the common ratio be $r$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

The formula for the sum of the first n terms of a G.P. is $S_n = \frac{a(r^n - 1)}{r - 1}$ (if $r \neq 1$).

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The sum of the first three terms is $S_3 = a_1 + a_2 + a_3 = a + ar + ar^2$.

Using the sum formula:

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The sum of the next three terms (4^th, 5^th, and 6^th) is $a_4 + a_5 + a_6$.

$a_4 = ar^{4-1} = ar^3$

$a_5 = ar^{5-1} = ar^4$

$a_6 = ar^{6-1} = ar^5$

The sum is $ar^3 + ar^4 + ar^5 = 128$.

Factor out $ar^3$ from this sum:

$ar^3(1 + r + r^2) = 128$.

Alternatively, we can express the sum of the next three terms as the sum of the first six terms minus the sum of the first three terms:

Sum of next three terms = $S_6 - S_3 = 128$.

The sum of the first six terms is $S_6 = \frac{a(r^6 - 1)}{r - 1}$.

So, $S_6 - S_3 = \frac{a(r^6 - 1)}{r - 1} - \frac{a(r^3 - 1)}{r - 1} = 128$

$\frac{a(r^6 - 1 - (r^3 - 1))}{r - 1} = 128$

$\frac{a(r^6 - r^3)}{r - 1} = 128$

Factor out $r^3$ from the numerator:

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Now we have a system of two equations (1) and (2):

Equation (1): $\frac{a(r^3 - 1)}{r - 1} = 16$

Equation (2): $\frac{ar^3(r^3 - 1)}{r - 1} = 128$

Substitute the value of $\frac{a(r^3 - 1)}{r - 1}$ from Equation (1) into Equation (2):

$r^3 \left(\frac{a(r^3 - 1)}{r - 1}\right) = 128$

$r^3 (16) = 128$

Solve for $r^3$:

$r^3 = \frac{128}{16}$

$r^3 = 8$

Take the cube root of both sides to find $r$:

$r = \sqrt[3]{8}$

$r = 2$

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Now that we have the common ratio $r=2$, substitute this value back into Equation (1) to find the first term $a$:

$\frac{a(2^3 - 1)}{2 - 1} = 16$

$\frac{a(8 - 1)}{1} = 16$

$\frac{a(7)}{1} = 16$

$7a = 16$

Solve for $a$:

$a = \frac{16}{7}$

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The first term is $a = \frac{16}{7}$ and the common ratio is $r = 2$.

Finally, find the sum to n terms ($S_n$) using the formula $S_n = \frac{a(r^n - 1)}{r - 1}$:

$S_n = \frac{\frac{16}{7}(2^n - 1)}{2 - 1}$

$S_n = \frac{\frac{16}{7}(2^n - 1)}{1}$

$S_n = \frac{16}{7}(2^n - 1)$

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Answer:

The first term of the G.P. is $a = \frac{16}{7}$.

The common ratio of the G.P. is $r = 2$.

The sum to n terms of the G.P. is $S_n = \frac{16}{7}(2^n - 1)$.

Given:

The first term of the G.P. is $a = 729$.

The 7^th term of the G.P. is $a_7 = 64$.

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To Find:

The sum of the first 7 terms of the G.P., $S_7$.

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Solution:

Let the first term of the G.P. be $a$ and the common ratio be $r$.

We are given $a = 729$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Using the given 7^th term ($n=7$):

$a_7 = ar^{7-1} = ar^6$

Substitute the values of $a_7$ and $a$:

$64 = 729 r^6$

Solve for $r^6$:

$r^6 = \frac{64}{729}$

To find $r$, take the 6^th root of both sides. Note that both 64 and 729 are perfect 6^th powers. $64 = 2^6$ and $729 = 3^6$.

$r^6 = \frac{2^6}{3^6} = \left(\frac{2}{3}\right)^6$

Taking the 6^th root, the common ratio can be positive or negative:

$r = \pm \frac{2}{3}$

---

We need to find the sum of the first 7 terms ($S_7$). The number of terms is $n=7$. The first term is $a=729$. The common ratio is $r$.

The formula for the sum of the first n terms of a G.P. is $S_n = \frac{a(1 - r^n)}{1 - r}$ (if $r \neq 1$).

Since both $r = \frac{2}{3}$ and $r = -\frac{2}{3}$ are not equal to 1, we use this formula.

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Case 1: Common ratio $r = \frac{2}{3}$

$S_7 = \frac{729 \left(1 - \left(\frac{2}{3}\right)^7\right)}{1 - \frac{2}{3}}$

$S_7 = \frac{729 \left(1 - \frac{2^7}{3^7}\right)}{\frac{1}{3}}$

Calculate $2^7 = 128$ and $3^7 = 2187$:

$S_7 = \frac{729 \left(1 - \frac{128}{2187}\right)}{\frac{1}{3}}$

$S_7 = 729 \times 3 \left(\frac{2187}{2187} - \frac{128}{2187}\right)$

$S_7 = 2187 \left(\frac{2187 - 128}{2187}\right)$

$S_7 = 2187 \left(\frac{2059}{2187}\right)$

$S_7 = 2059$

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Case 2: Common ratio $r = -\frac{2}{3}$

$S_7 = \frac{729 \left(1 - \left(-\frac{2}{3}\right)^7\right)}{1 - \left(-\frac{2}{3}\right)}$

Calculate $\left(-\frac{2}{3}\right)^7 = \frac{(-2)^7}{3^7} = \frac{-128}{2187} = -\frac{128}{2187}$.

The numerator is $1 - \left(-\frac{128}{2187}\right) = 1 + \frac{128}{2187} = \frac{2187 + 128}{2187} = \frac{2315}{2187}$.

The denominator is $1 - \left(-\frac{2}{3}\right) = 1 + \frac{2}{3} = \frac{3}{3} + \frac{2}{3} = \frac{5}{3}$.

$S_7 = \frac{729 \left(\frac{2315}{2187}\right)}{\frac{5}{3}}$

$S_7 = 729 \times \frac{2315}{2187} \times \frac{3}{5}$

Note that $2187 = 3^7$ and $729 = 3^6$.

$S_7 = 3^6 \times \frac{2315}{3^7} \times \frac{3}{5}$

$S_7 = \frac{3^6 \times 2315 \times 3}{3^7 \times 5}$

$S_7 = \frac{3^7 \times 2315}{3^7 \times 5}$

$S_7 = \frac{2315}{5}$

$S_7 = 463$

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The problem does not specify that the terms are positive, so both values of the common ratio are possible, leading to two possible sums for the first 7 terms.

---

Answer:

The common ratio is $\frac{2}{3}$ or $-\frac{2}{3}$.

The sum of the first 7 terms is 2059 (if $r = \frac{2}{3}$) or 463 (if $r = -\frac{2}{3}$).

Given:

The sum of the first two terms of a G.P. is $-4$.

The fifth term of the G.P. is 4 times the third term.

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To Find:

The geometric progression (G.P.), which means finding the first term and the common ratio.

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Solution:

Let the first term of the G.P. be $a$ and the common ratio be $r$.

The terms of the G.P. are $a, ar, ar^2, ar^3, ar^4, \dots, ar^{n-1}, \dots$

The formula for the n^th term is $a_n = ar^{n-1}$.

---

According to the first condition, the sum of the first two terms is $-4$:

$a_1 + a_2 = -4$

$a + ar = -4$

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According to the second condition, the fifth term is 4 times the third term:

$a_5 = 4 \times a_3$

Using the formula for the n^th term:

$ar^{5-1} = 4 \times ar^{3-1}$

$ar^4 = 4ar^2$

Assuming the G.P. is non-trivial ($a \neq 0$ and $r \neq 0$), we can divide both sides by $ar^2$:

$\frac{ar^4}{ar^2} = \frac{4ar^2}{ar^2}$

$r^{4-2} = 4$

Taking the square root of both sides of Equation (2):

$r = \pm \sqrt{4}$

$r = \pm 2$

So, there are two possible values for the common ratio: $r = 2$ or $r = -2$.

---

Now, substitute these values of $r$ back into Equation (1) to find the corresponding value(s) of $a$.

Equation (1) is $a(1 + r) = -4$.

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Case 1: Common ratio $r = 2$

Substitute $r=2$ into Equation (1):

$a(1 + 2) = -4$

$a(3) = -4$

$a = -\frac{4}{3}$

In this case, the first term is $a = -\frac{4}{3}$ and the common ratio is $r = 2$. The G.P. starts as $-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots$

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Case 2: Common ratio $r = -2$

Substitute $r=-2$ into Equation (1):

$a(1 + (-2)) = -4$

$a(-1) = -4$

$a = 4$

In this case, the first term is $a = 4$ and the common ratio is $r = -2$. The G.P. starts as $4, -8, 16, \dots$

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Both cases satisfy the given conditions. The problem asks for "a G.P.", implying at least one solution, and we found two.

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Answer:

There are two possible geometric progressions that satisfy the given conditions:

1. First term $a = -\frac{4}{3}$ and common ratio $r = 2$. The G.P. is $-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, -\frac{32}{3}, -\frac{64}{3}, \dots$

2. First term $a = 4$ and common ratio $r = -2$. The G.P. is $4, -8, 16, -32, 64, \dots$

Given:

The 4^th term of a G.P. is $a_4 = x$.

The 10^th term of the same G.P. is $a_{10} = y$.

The 16^th term of the same G.P. is $a_{16} = z$.

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To Prove:

x, y, z are in G.P.

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Proof:

Let the first term of the G.P. be $a$ and the common ratio be $r$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Using this formula, we can express the given terms:

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For three numbers x, y, and z to be in G.P., the ratio of consecutive terms must be the same. That is, $\frac{y}{x} = \frac{z}{y}$, or equivalently, $y^2 = xz$ (assuming $x \neq 0$ and $z \neq 0$).

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Consider the square of the middle term, $y^2$, from Equation (2):

$y = ar^9$

$y^2 = (ar^9)^2$

$y^2 = a^2 (r^9)^2$

$y^2 = a^2 r^{18}$

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Consider the product of the first and third terms, $xz$, from Equations (1) and (3):

$xz = (ar^3)(ar^{15})$

$xz = a \cdot a \cdot r^3 \cdot r^{15}$

$xz = a^{1+1} r^{3+15}$

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Comparing Equation (4) and Equation (5), we see that $y^2 = a^2 r^{18}$ and $xz = a^2 r^{18}$.

Therefore, $y^2 = xz$.

This is the condition for three numbers x, y, and z to be in G.P.

(Assuming $a \neq 0$ and $r \neq 0$, which are usually implicit for a G.P. unless otherwise specified. If $a=0$, all terms are 0, and $0, 0, 0$ is a G.P. If $r=1$, the sequence is $a, a, a$. If $a \neq 0$, this is a G.P. with $r=1$. If $r \neq 0, 1$ and $a \neq 0$, $x, y, z$ are non-zero, and $y^2=xz$ implies $\frac{y}{x} = \frac{z}{y}$).

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Answer:

Hence, it is proved that x, y, z are in G.P.

Given:

The sequence is 8, 88, 888, 8888, ... to n terms.

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To Find:

The sum of the first n terms of the given sequence.

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Solution:

Let the sum of the first n terms be $S_n$.

The terms of the sequence are $a_1 = 8$, $a_2 = 88$, $a_3 = 888$, and so on.

The n^th term, $a_n$, is a number consisting of n digits, all equal to 8.

We can write each term by factoring out 8:

$a_1 = 8 = 8 \times 1$

$a_2 = 88 = 8 \times 11$

$a_3 = 888 = 8 \times 111$

$a_n = 8 \times \underbrace{11\dots1}_{\text{n times}}$

The number $\underbrace{11\dots1}_{\text{n times}}$ can be expressed as $\frac{10^n - 1}{9}$.

So, the n^th term is $a_n = 8 \times \frac{10^n - 1}{9} = \frac{8}{9}(10^n - 1)$.

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The sum of the first n terms is $S_n = a_1 + a_2 + \dots + a_n = \sum\limits_{k=1}^{n} a_k$.

$S_n = \sum\limits_{k=1}^{n} \frac{8}{9}(10^k - 1)$

$S_n = \frac{8}{9} \sum\limits_{k=1}^{n} (10^k - 1)$

Separate the summation:

$S_n = \frac{8}{9} \left( \sum\limits_{k=1}^{n} 10^k - \sum\limits_{k=1}^{n} 1 \right)$

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The second summation is simply the sum of n ones:

$\sum\limits_{k=1}^{n} 1 = 1 + 1 + \dots + 1$ (n times) $= n$.

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The first summation is the sum of a geometric series:

$\sum\limits_{k=1}^{n} 10^k = 10^1 + 10^2 + 10^3 + \dots + 10^n$.

This is a geometric progression with the first term $A = 10$, the common ratio $R = 10$, and the number of terms is $n$.

The sum of the first n terms of a G.P. is $S_n = \frac{A(R^n - 1)}{R - 1}$.

Sum of this G.P. $= \frac{10(10^n - 1)}{10 - 1} = \frac{10(10^n - 1)}{9}$.

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Substitute these results back into the expression for $S_n$:

$S_n = \frac{8}{9} \left( \frac{10(10^n - 1)}{9} - n \right)$

This can be further simplified:

$S_n = \frac{8}{9} \times \frac{10(10^n - 1)}{9} - \frac{8}{9}n$

$S_n = \frac{80(10^n - 1)}{81} - \frac{8n}{9}$

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Answer:

The sum of the first n terms of the sequence is $S_n = \frac{80(10^n - 1)}{81} - \frac{8n}{9}$.

Given:

Sequence 1: $A = \{2, 4, 8, 16, 32\}$

Sequence 2: $B = \{128, 32, 8, 2, \frac{1}{2}\}$

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To Find:

The sum of the products of the corresponding terms of the two sequences.

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Solution:

Let the terms of the first sequence be $a_k$ and the terms of the second sequence be $b_k$, for $k=1, 2, 3, 4, 5$.

The first sequence is 2, 4, 8, 16, 32.

This is a G.P. with first term $a = 2$ and common ratio $r_A = \frac{4}{2} = 2$.

The k^th term of the first sequence is $a_k = 2 \cdot 2^{k-1} = 2^k$.

---

The second sequence is 128, 32, 8, 2, $\frac{1}{2}$.

This is a G.P. with first term $b = 128$ and common ratio $r_B = \frac{32}{128} = \frac{1}{4}$.

The k^th term of the second sequence is $b_k = 128 \cdot \left(\frac{1}{4}\right)^{k-1}$.

---

We need to find the sum of the products of the corresponding terms, which is $\sum\limits_{k=1}^{5} a_k b_k$.

The k^th product term is $p_k = a_k \times b_k$.

$p_k = (2^k) \times \left(128 \cdot \left(\frac{1}{4}\right)^{k-1}\right)$

Express 128 as a power of 2: $128 = 2^7$.

Express $\frac{1}{4}$ as a power of 2: $\frac{1}{4} = 4^{-1} = (2^2)^{-1} = 2^{-2}$.

Substitute these into the expression for $p_k$:

$p_k = 2^k \times 2^7 \times (2^{-2})^{k-1}$

$p_k = 2^{k+7} \times 2^{-2(k-1)}$

$p_k = 2^{k+7} \times 2^{-2k+2}$

Using the exponent rule $x^m \cdot x^n = x^{m+n}$:

$p_k = 2^{(k+7) + (-2k+2)}$

$p_k = 2^{k+7-2k+2}$

$p_k = 2^{9-k}$

---

The sequence of products is $p_k = 2^{9-k}$ for $k=1, 2, 3, 4, 5$.

The terms of this product sequence are:

$p_1 = 2^{9-1} = 2^8 = 256$

$p_2 = 2^{9-2} = 2^7 = 128$

$p_3 = 2^{9-3} = 2^6 = 64$

$p_4 = 2^{9-4} = 2^5 = 32$

$p_5 = 2^{9-5} = 2^4 = 16$

The sequence of products is 256, 128, 64, 32, 16.

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This sequence of products is a G.P. with first term $P = 256$ and common ratio $R = \frac{128}{256} = \frac{1}{2}$. The number of terms is $N=5$.

The sum of this G.P. is given by $S_N = \frac{P(1 - R^N)}{1 - R}$, since $|R| < 1$.

Sum $= \frac{256 \left(1 - \left(\frac{1}{2}\right)^5\right)}{1 - \frac{1}{2}}$

Sum $= \frac{256 \left(1 - \frac{1}{32}\right)}{\frac{1}{2}}$

Sum $= 256 \times 2 \times \left(\frac{32 - 1}{32}\right)$

Sum $= 512 \times \frac{31}{32}$

Cancel out the common factor:

Sum $= \frac{\cancel{512}^{16} \times 31}{\cancel{32}_{1}}$

Sum $= 16 \times 31$

Perform the multiplication:

Sum $= 496$

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Answer:

The sum of the products of the corresponding terms is 496.

Given:

Two sequences which are geometric progressions:

Sequence 1: a, ar, ar², ..., ar^n-1

Sequence 2: A, AR, AR², ..., AR^n-1

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To Show:

The sequence formed by the products of the corresponding terms is a G.P.

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To Find:

The common ratio of the resulting G.P.

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Proof:

Let the terms of the first sequence be $a_k$ for $k=1, 2, \dots, n$. The first term is $a$ and the common ratio is $r$. The k^th term is $a_k = ar^{k-1}$.

Let the terms of the second sequence be $b_k$ for $k=1, 2, \dots, n$. The first term is $A$ and the common ratio is $R$. The k^th term is $b_k = AR^{k-1}$.

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Consider the sequence formed by the products of the corresponding terms. Let the terms of this new sequence be $c_k = a_k \times b_k$.

The k^th term of the product sequence is:

$c_k = a_k \times b_k = (ar^{k-1}) \times (AR^{k-1})$

$c_k = (aA) (rR)^{k-1}$

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To show that this sequence is a G.P., we need to show that the ratio of any term to its preceding term is a constant value.

Consider the ratio of the (k+1)^th term to the k^th term:

$\frac{c_{k+1}}{c_k} = \frac{(aA)(rR)^{(k+1)-1}}{(aA)(rR)^{k-1}}$

$\frac{c_{k+1}}{c_k} = \frac{(aA)(rR)^k}{(aA)(rR)^{k-1}}$

Assuming $aA \neq 0$ (for a non-trivial product sequence), we can cancel $(aA)$ from the numerator and denominator:

$\frac{c_{k+1}}{c_k} = \frac{(rR)^k}{(rR)^{k-1}}$

Using the exponent rule $\frac{x^m}{x^n} = x^{m-n}$:

$\frac{c_{k+1}}{c_k} = (rR)^{k - (k-1)}$

$\frac{c_{k+1}}{c_k} = (rR)^{k - k + 1}$

$\frac{c_{k+1}}{c_k} = (rR)^1 = rR$

---

The ratio of consecutive terms in the sequence $\{c_k\}$ is $rR$, which is a constant value (since $r$ and $R$ are constants). This means the sequence of products is a geometric progression.

The first term of this product G.P. is $c_1 = a_1 b_1 = aA$.

The common ratio of this product G.P. is the constant ratio we found, which is $rR$.

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Answer:

The sequence formed by the products of the corresponding terms is $(aA)(rR)^0, (aA)(rR)^1, \dots, (aA)(rR)^{n-1}$, which is a G.P.

The common ratio of this G.P. is $rR$.

Given:

We are looking for four numbers that form a geometric progression (G.P.).

Let the four terms of the G.P. be $a_1, a_2, a_3, a_4$. Let the first term be $a$ and the common ratio be $r$. The terms are $a, ar, ar^2, ar^3$.

Condition 1: The third term is greater than the first term by 9. ($a_3 = a_1 + 9$).

Condition 2: The second term is greater than the 4^th term by 18. ($a_2 = a_4 + 18$).

---

To Find:

The four numbers forming the G.P.

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Solution:

Using the terms in terms of $a$ and $r$, we write the given conditions as equations:

Condition 1: $a_3 = a_1 + 9$

$ar^2 = a + 9$

---

Condition 2: $a_2 = a_4 + 18$

$ar = ar^3 + 18$

---

We have a system of two equations with two variables $a$ and $r$:

(1) $a(r^2 - 1) = 9$

(2) $ar(1 - r^2) = 18$

From Equation (1), we can write $a(r^2 - 1) = 9$.

From Equation (2), we can rewrite $ar(1 - r^2) = ar(-(r^2 - 1)) = -ar(r^2 - 1) = 18$.

So, $-ar(r^2 - 1) = 18$.

Substitute the value of $a(r^2 - 1)$ from Equation (1) into this equation:

$-r \times [a(r^2 - 1)] = 18$

$-r \times [9] = 18$

$-9r = 18$

Divide both sides by -9:

$r = \frac{18}{-9}$

$r = -2$

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Now that we have the common ratio $r = -2$, substitute this value back into Equation (1) to find the first term $a$:

$a(r^2 - 1) = 9$

$a((-2)^2 - 1) = 9$

$a(4 - 1) = 9$

$a(3) = 9$

Divide both sides by 3:

$a = 3$

---

The first term is $a = 3$ and the common ratio is $r = -2$.

The four numbers forming the G.P. are:

$a_1 = a = 3$

$a_2 = ar = 3 \times (-2) = -6$

$a_3 = ar^2 = 3 \times (-2)^2 = 3 \times 4 = 12$

$a_4 = ar^3 = 3 \times (-2)^3 = 3 \times (-8) = -24$

The four numbers are 3, -6, 12, -24.

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Verify the given conditions:

Condition 1: The third term is greater than the first term by 9.

$a_3 - a_1 = 12 - 3 = 9$. (Correct)

Condition 2: The second term is greater than the 4^th term by 18.

$a_2 - a_4 = -6 - (-24) = -6 + 24 = 18$. (Correct)

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Answer:

The four numbers forming the geometric progression are 3, -6, 12, and -24.

Given:

Let the first term of the Geometric Progression (G.P.) be $A$ and the common ratio be $R$.

The $p^{\text{th}}$ term of the G.P. is $a$. Using the formula for the $n^{\text{th}}$ term of a G.P., $a_n = AR^{n-1}$, we have:

The $q^{\text{th}}$ term of the G.P. is $b$.

The $r^{\text{th}}$ term of the G.P. is $c$.

---

To Prove:

We need to prove that $a^{q-r} b^{r-p} c^{p-q} = 1$.

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Proof:

Consider the Left-Hand Side (LHS) of the expression:

$LHS = a^{q-r} b^{r-p} c^{p-q}$

Substitute the values of $a, b,$ and $c$ from equations (i), (ii), and (iii):

$LHS = (AR^{p-1})^{q-r} \cdot (AR^{q-1})^{r-p} \cdot (AR^{r-1})^{p-q}$

Using the exponent rule $(xy)^m = x^m y^m$, we separate the terms with base $A$ and base $R$:

$LHS = (A^{q-r} \cdot (R^{p-1})^{q-r}) \cdot (A^{r-p} \cdot (R^{q-1})^{r-p}) \cdot (A^{p-q} \cdot (R^{r-1})^{p-q})$

Group the terms with base $A$ and base $R$ together:

$LHS = (A^{q-r} \cdot A^{r-p} \cdot A^{p-q}) \cdot (R^{(p-1)(q-r)} \cdot R^{(q-1)(r-p)} \cdot R^{(r-1)(p-q)})$

Using the exponent rule $x^m \cdot x^n = x^{m+n}$, we add the exponents for each base:

$LHS = A^{(q-r) + (r-p) + (p-q)} \cdot R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)}$

Now, let's simplify the exponents for $A$ and $R$ separately.

Exponent of A:

$(q-r) + (r-p) + (p-q) = q - r + r - p + p - q \ $$ = (q-q) + (-r+r) + (-p+p) = 0$.

Exponent of R:

$(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)$

Expanding each term:

$= (pq - pr - q + r) + (qr - pq - r + p) + (pr - qr - p + q)$

Combining the like terms:

$= (pq - pq) + (-pr + pr) + (-q + q) + (r - r) + (qr - qr) + (p - p)$

$= 0 + 0 + 0 + 0 + 0 + 0 = 0$.

Now substitute the simplified exponents back into the expression for LHS:

$LHS = A^0 \cdot R^0$

Since any non-zero number raised to the power of 0 is 1 (assuming $A \neq 0, R \neq 0$):

$LHS = 1 \cdot 1 = 1$.

Thus, $LHS = 1 = RHS$ (Right-Hand Side).

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Hence, Proved.

Given:

First term of a G.P. is $a_1=a$.

$n^{\text{th}}$ term is $a_n=b$.

$P$ is the product of the first $n$ terms.

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To Prove:

$P^2 = (ab)^n$.

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Proof:

Let the common ratio be $r$. The terms of the G.P. are $a, ar, ar^2, \dots, ar^{n-1}$.

The product $P = a \cdot ar \cdot ar^2 \cdots ar^{n-1} = a^n r^{0+1+2+\dots+(n-1)}$.

The sum of the exponents of $r$ is $\frac{(n-1)n}{2}$.

So, $P = a^n r^{\frac{n(n-1)}{2}}$.

Squaring both sides: $P^2 = (a^n)^2 (r^{\frac{n(n-1)}{2}})^2 = a^{2n} r^{n(n-1)}$.

Now consider the RHS: $(ab)^n$.

We are given $b = a_n = ar^{n-1}$.

RHS = $(a \cdot (ar^{n-1}))^n = (a^2 r^{n-1})^n = (a^2)^n (r^{n-1})^n = a^{2n} r^{n(n-1)}$.

Since LHS = $a^{2n} r^{n(n-1)}$ and RHS = $a^{2n} r^{n(n-1)}$, we have $P^2 = (ab)^n$.

Hence, proved.

To Show:

$\frac{\text{Sum of first n terms}}{\text{Sum of terms from (n+1) to 2n}} = \frac{1}{r^n}$.

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Proof:

Let the G.P. have first term $a$ and common ratio $r$.

Sum of first $n$ terms, $S_n = \frac{a(r^n-1)}{r-1}$.

The sum of terms from $(n+1)$ to $2n$ is the sum of the first $2n$ terms minus the sum of the first $n$ terms: $S_{2n} - S_n$.

$S_{2n} = \frac{a(r^{2n}-1)}{r-1}$.

$S_{2n} - S_n = \frac{a(r^{2n}-1)}{r-1} - \frac{a(r^n-1)}{r-1} = \frac{a}{r-1}[(r^{2n}-1) - (r^n-1)] \ $$ = \frac{a}{r-1}[r^{2n}-r^n] = \frac{a r^n(r^n-1)}{r-1}$.

Now, find the required ratio:

$\frac{S_n}{S_{2n} - S_n} = \frac{\frac{a(r^n-1)}{r-1}}{\frac{ar^n(r^n-1)}{r-1}}$.

Canceling the common factors $\frac{a(r^n-1)}{r-1}$, we get:

Ratio = $\frac{1}{r^n}$.

Hence, proved.

Given:

$a, b, c, d$ are in G.P. Let the common ratio be $r$.

Then $b=ar, c=ar^2, d=ar^3$.

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To Show:

$(a^2+b^2+c^2)(b^2+c^2+d^2) = (ab+bc+cd)^2$.

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Proof:

LHS = $(a^2+b^2+c^2)(b^2+c^2+d^2)$

$= (a^2 + (ar)^2 + (ar^2)^2)((ar)^2 + (ar^2)^2 + (ar^3)^2)$

$= (a^2(1+r^2+r^4))(a^2r^2(1+r^2+r^4))$

$= a^4r^2(1+r^2+r^4)^2$.

RHS = $(ab+bc+cd)^2$

$= (a(ar) + (ar)(ar^2) + (ar^2)(ar^3))^2$

$= (a^2r + a^2r^3 + a^2r^5)^2$

$= (a^2r(1+r^2+r^4))^2$

$= (a^2r)^2(1+r^2+r^4)^2 = a^4r^2(1+r^2+r^4)^2$.

Since LHS = RHS, the statement is proved.

Given:

We need to insert 2 numbers between 3 and 81 to form a G.P.

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Solution:

Let the two numbers be $g_1, g_2$. The G.P. is $3, g_1, g_2, 81$.

First term, $a=3$. Total terms, $n=4$. The 4th term, $a_4=81$.

Using the formula $a_n = ar^{n-1}$:

$a_4 = ar^3 \implies 81 = 3 \cdot r^3 \implies r^3 = 27 \implies r=3$.

The common ratio is 3. The two numbers are:

$g_1 = ar = 3(3)=9$.

$g_2 = ar^2 = 3(3^2)=27$.

The two numbers are 9 and 27.

Given:

The expression $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is the G.M. between $a$ and $b$.

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Solution:

The G.M. of $a$ and $b$ is $\sqrt{ab}$.

$\frac{a^{n+1}+b^{n+1}}{a^n+b^n} = \sqrt{ab} = a^{1/2}b^{1/2}$.

$a^{n+1}+b^{n+1} = (a^n+b^n)a^{1/2}b^{1/2} = a^{n+1/2}b^{1/2} + a^{1/2}b^{n+1/2}$.

$a^{n+1} - a^{n+1/2}b^{1/2} = a^{1/2}b^{n+1/2} - b^{n+1}$.

$a^{n+1/2}(a^{1/2} - b^{1/2}) = b^{n+1/2}(a^{1/2} - b^{1/2})$.

Assuming $a \neq b$, we can divide by $(a^{1/2} - b^{1/2})$.

$a^{n+1/2} = b^{n+1/2} \implies \left(\frac{a}{b}\right)^{n+1/2} = 1 = \left(\frac{a}{b}\right)^0$.

Equating exponents: $n+\frac{1}{2}=0 \implies n = -\frac{1}{2}$.

The value of $n$ is -1/2.

Given:

For two numbers $a$ and $b$, $a+b = 6\sqrt{ab}$.

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To Show:

The numbers are in the ratio $(3+2\sqrt{2}) : (3-2\sqrt{2})$.

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Proof:

$a+b = 6\sqrt{ab} \implies \frac{a+b}{2\sqrt{ab}} = 3$.

Using Componendo and Dividendo:

$\frac{(a+b)+2\sqrt{ab}}{(a+b)-2\sqrt{ab}} = \frac{3+1}{3-1} = \frac{4}{2} = 2$.

The LHS is $\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2} = \left(\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right)^2 = 2$.

$\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \sqrt{2}$.

Again using Componendo and Dividendo:

$\frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}-\sqrt{b})} = \frac{\sqrt{2}+1}{\sqrt{2}-1}$.

$\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1} \implies \frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1}$.

Squaring both sides: $\frac{a}{b} = \frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2} = \frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}} = \frac{3+2\sqrt{2}}{3-2\sqrt{2}}$.

Hence, proved.

Given:

For two numbers $a, b$, A.M. is $A = \frac{a+b}{2}$ and G.M. is $G = \sqrt{ab}$.

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To Prove:

The numbers are $A \pm \sqrt{(A+G)(A-G)}$.

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Proof:

The quadratic equation with roots $a, b$ is $x^2 - (a+b)x + ab = 0$.

From the given relations, $a+b = 2A$ and $ab=G^2$.

The equation is $x^2 - 2Ax + G^2 = 0$.

The roots of this equation are given by the quadratic formula:

$x = \frac{-(-2A) \pm \sqrt{(-2A)^2 - 4(1)(G^2)}}{2(1)} = \frac{2A \pm \sqrt{4A^2 - 4G^2}}{2} = \frac{2A \pm 2\sqrt{A^2-G^2}}{2}$.

$x = A \pm \sqrt{A^2 - G^2}$.

Using the difference of squares, $A^2 - G^2 = (A+G)(A-G)$.

So, the numbers are $A \pm \sqrt{(A+G)(A-G)}$.

Hence, proved.

Given:

The number of bacteria doubles every hour, which forms a G.P.

Original number of bacteria (initial term), $a_0 = 30$.

Common ratio, $r=2$.

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Solution:

Let $N(t)$ be the number of bacteria at the end of hour $t$.

$N(t) = a_0 \cdot r^t = 30 \cdot 2^t$.

At the end of the 2nd hour ($t=2$):

$N(2) = 30 \cdot 2^2 = 30 \cdot 4 = 120$.

At the end of the 4th hour ($t=4$):

$N(4) = 30 \cdot 2^4 = 30 \cdot 16 = 480$.

At the end of the n-th hour ($t=n$):

$N(n) = 30 \cdot 2^n$.

The number of bacteria will be 120 at the end of the 2nd hour, 480 at the end of the 4th hour, and $30 \cdot 2^n$ at the end of the nth hour.

Given:

Principal amount, $P = \textsf{₹} 500$.

Annual interest rate, $R = 10\% = 0.1$.

Time period, $n=10$ years.

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Solution:

We use the formula for compound interest: $A = P(1+R)^n$.

$A = 500(1+0.1)^{10} = 500(1.1)^{10}$.

Using a calculator, $(1.1)^{10} \approx 2.59374$.

$A \approx 500 \times 2.59374 = 1296.87$.

The amount will be approximately $\textsf{₹} 1296.87$.

Given:

For the roots of a quadratic equation, let them be $\alpha, \beta$.

A.M. = $\frac{\alpha+\beta}{2} = 8 \implies \alpha+\beta = 16$.

G.M. = $\sqrt{\alpha\beta} = 5 \implies \alpha\beta = 25$.

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Solution:

A quadratic equation with roots $\alpha, \beta$ is given by:

$x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0$.

$x^2 - (\alpha+\beta)x + (\alpha\beta) = 0$.

Substitute the values we found:

$x^2 - 16x + 25 = 0$.

The quadratic equation is $x^2 - 16x + 25 = 0$.

Given:

The series is $S_n = 5 + 11 + 19 + 29 + 41 + \dots$ to $n$ terms.

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To Find:

The sum of the series to $n$ terms, $S_n$.

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Solution:

Let the $n^{\text{th}}$ term be $a_n$. We find the differences between consecutive terms:

$a_2 - a_1 = 11 - 5 = 6$

$a_3 - a_2 = 19 - 11 = 8$

$a_4 - a_3 = 29 - 19 = 10$

The differences (6, 8, 10, ...) are in an Arithmetic Progression (A.P.) with a first term of 6 and a common difference of 2. This means the original series is a quadratic sequence.

The general term $a_n$ is a quadratic of the form $An^2+Bn+C$. By solving for the first few terms, we find the formula for the $n^{\text{th}}$ term is $a_n = n^2 + 3n + 1$.

Now, we need to find the sum of this sequence:

$S_n = \sum\limits_{k=1}^{n} a_k = \sum\limits_{k=1}^{n} (k^2 + 3k + 1)$

$S_n = \sum\limits_{k=1}^{n} k^2 + 3\sum\limits_{k=1}^{n} k + \sum\limits_{k=1}^{n} 1$

Using the standard sum formulas:

$\sum\limits k^2 = \frac{n(n+1)(2n+1)}{6}$, $\sum\limits k = \frac{n(n+1)}{2}$, $\sum\limits 1 = n$.

$S_n = \frac{n(n+1)(2n+1)}{6} + 3\frac{n(n+1)}{2} + n$

Taking $n$ as a common factor:

$S_n = n \left[ \frac{(n+1)(2n+1)}{6} + \frac{3(n+1)}{2} + 1 \right]$

$S_n = n \left[ \frac{2n^2+3n+1 + 9(n+1) + 6}{6} \right]$

$S_n = n \left[ \frac{2n^2+3n+1+9n+9+6}{6} \right] = \frac{n}{6}(2n^2+12n+16)$

$S_n = \frac{2n}{6}(n^2+6n+8) = \frac{n}{3}(n+2)(n+4)$.

The sum to n terms is $\frac{n(n+2)(n+4)}{3}$.

Given:

The $n^{\text{th}}$ term of a series is $a_n = n(n+3) = n^2 + 3n$.

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To Find:

The sum of the series to $n$ terms, $S_n$.

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Solution:

The sum to $n$ terms is the summation of the $k^{\text{th}}$ term from $k=1$ to $n$.

$S_n = \sum\limits_{k=1}^{n} a_k = \sum\limits_{k=1}^{n} (k^2 + 3k)$

$S_n = \sum\limits_{k=1}^{n} k^2 + 3\sum\limits_{k=1}^{n} k$

Using the standard sum formulas:

$\sum\limits k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum\limits k = \frac{n(n+1)}{2}$.

$S_n = \frac{n(n+1)(2n+1)}{6} + 3\frac{n(n+1)}{2}$

Factor out the common term $\frac{n(n+1)}{2}$:

$S_n = \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 3 \right] = \frac{n(n+1)}{2} \left[ \frac{2n+1+9}{3} \right]$

$S_n = \frac{n(n+1)}{2} \left[ \frac{2n+10}{3} \right] = \frac{n(n+1)}{2} \cdot \frac{2(n+5)}{3}$

$S_n = \frac{n(n+1)(n+5)}{3}$.

The sum to n terms is $\frac{n(n+1)(n+5)}{3}$.

To Find:

The sum to $n$ terms of the series $1 \times 2 + 2 \times 3 + 3 \times 4 + \dots$

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Solution:

The $k^{\text{th}}$ term of the series is $a_k = k(k+1) = k^2 + k$.

The sum to $n$ terms is $S_n = \sum\limits_{k=1}^{n} a_k = \sum\limits_{k=1}^{n} (k^2 + k) = \sum\limits_{k=1}^{n} k^2 + \sum\limits_{k=1}^{n} k$.

Using the formulas for the sum of the first $n$ squares and the first $n$ natural numbers:

$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$.

Factor out the common term $\frac{n(n+1)}{2}$:

$S_n = \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 1 \right] = \frac{n(n+1)}{2} \left[ \frac{2n+1+3}{3} \right]$.

$S_n = \frac{n(n+1)}{2} \left[ \frac{2n+4}{3} \right] = \frac{n(n+1) \cdot 2(n+2)}{6} = \frac{n(n+1)(n+2)}{3}$.

The sum is $\frac{n(n+1)(n+2)}{3}$.

To Find:

The sum to $n$ terms of the series $1 \times 2 \times 3 + 2 \times 3 \times 4 + \dots$

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Solution:

The $k^{\text{th}}$ term is $a_k = k(k+1)(k+2) = k^3 + 3k^2 + 2k$.

$S_n = \sum\limits_{k=1}^{n} (k^3 + 3k^2 + 2k) = \sum\limits k^3 + 3\sum\limits k^2 + 2\sum\limits k$.

$S_n = \left(\frac{n(n+1)}{2}\right)^2 + 3\frac{n(n+1)(2n+1)}{6} + 2\frac{n(n+1)}{2}$.

$S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2} + n(n+1)$.

Factor out $\frac{n(n+1)}{4}$:

$S_n = \frac{n(n+1)}{4} [n(n+1) + 2(2n+1) + 4] = \frac{n(n+1)}{4} [n^2+n+4n+2+4]$.

$S_n = \frac{n(n+1)}{4} [n^2+5n+6] = \frac{n(n+1)(n+2)(n+3)}{4}$.

The sum is $\frac{n(n+1)(n+2)(n+3)}{4}$.

To Find:

The sum to $n$ terms of the series $3 \times 1^2 + 5 \times 2^2 + \dots$

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Solution:

The $k^{\text{th}}$ term has two parts. The first part is an A.P. $3, 5, 7, \dots$ whose $k^{\text{th}}$ term is $3+(k-1)2 = 2k+1$. The second part is $k^2$.

The $k^{\text{th}}$ term of the series is $a_k = (2k+1)k^2 = 2k^3 + k^2$.

$S_n = \sum\limits_{k=1}^{n} (2k^3 + k^2) = 2\sum\limits k^3 + \sum\limits k^2$.

$S_n = 2\left(\frac{n(n+1)}{2}\right)^2 + \frac{n(n+1)(2n+1)}{6} = \frac{n^2(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{6}$.

Factor out $\frac{n(n+1)}{6}$:

$S_n = \frac{n(n+1)}{6} [3n(n+1) + (2n+1)] = \frac{n(n+1)}{6} [3n^2+3n+2n+1]$.

$S_n = \frac{n(n+1)(3n^2+5n+1)}{6}$.

The sum is $\frac{n(n+1)(3n^2+5n+1)}{6}$.

To Find:

The sum to $n$ terms of the series $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \dots$

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Solution:

The $k^{\text{th}}$ term is $a_k = \frac{1}{k(k+1)}$.

Using partial fraction decomposition, we can write $a_k = \frac{1}{k} - \frac{1}{k+1}$.

The sum $S_n$ is a telescoping series:

$S_n = \sum\limits_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{k+1}\right)$.

$S_n = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$.

All intermediate terms cancel out, leaving:

$S_n = 1 - \frac{1}{n+1} = \frac{n+1-1}{n+1} = \frac{n}{n+1}$.

The sum is $\frac{n}{n+1}$.

To Find:

The sum of the series $5^2 + 6^2 + \dots + 20^2$.

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Solution:

This is not a sum starting from 1. We can write it as the difference of two sums:

Sum = $(\sum\limits_{k=1}^{20} k^2) - (\sum\limits_{k=1}^{4} k^2)$.

Using the formula $\sum\limits k^2 = \frac{n(n+1)(2n+1)}{6}$:

$\sum\limits_{k=1}^{20} k^2 = \frac{20(21)(41)}{6} = 10 \times 7 \times 41 = 2870$.

$\sum\limits_{k=1}^{4} k^2 = \frac{4(5)(9)}{6} = 2 \times 5 \times 3 = 30$.

The required sum = $2870 - 30 = 2840$.

The sum of the series is 2840.

To Find:

The sum to $n$ terms of the series $3 \times 8 + 6 \times 11 + \dots$

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Solution:

The $k^{\text{th}}$ term is a product of two terms from two different A.P.s.

First A.P.: 3, 6, 9, ... whose $k^{\text{th}}$ term is $3k$.

Second A.P.: 8, 11, 14, ... whose $k^{\text{th}}$ term is $8+(k-1)3 = 3k+5$.

The $k^{\text{th}}$ term of the series is $a_k = (3k)(3k+5) = 9k^2 + 15k$.

$S_n = \sum\limits_{k=1}^{n} (9k^2 + 15k) = 9\sum\limits k^2 + 15\sum\limits k$.

$S_n = 9\frac{n(n+1)(2n+1)}{6} + 15\frac{n(n+1)}{2} = \frac{3n(n+1)(2n+1)}{2} + \frac{15n(n+1)}{2}$.

Factor out $\frac{3n(n+1)}{2}$:

$S_n = \frac{3n(n+1)}{2} [(2n+1) + 5] = \frac{3n(n+1)}{2} [2n+6] = \frac{3n(n+1) \cdot 2(n+3)}{2}$.

$S_n = 3n(n+1)(n+3)$.

The sum is $3n(n+1)(n+3)$.

To Find:

The sum to $n$ terms of the given series.

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Solution:

The $k^{\text{th}}$ term of the series, $a_k$, is the sum of the squares of the first $k$ natural numbers.

$a_k = \sum\limits_{i=1}^{k} i^2 = \frac{k(k+1)(2k+1)}{6} = \frac{1}{6}(2k^3 + 3k^2 + k)$.

The sum to $n$ terms is $S_n = \sum\limits_{k=1}^{n} a_k = \frac{1}{6} \sum\limits_{k=1}^{n} (2k^3 + 3k^2 + k)$.

$S_n = \frac{1}{6} \left[ 2\sum\limits k^3 + 3\sum\limits k^2 + \sum\limits k \right]$.

$S_n = \frac{1}{6} \left[ 2\frac{n^2(n+1)^2}{4} + 3\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right]$.

$S_n = \frac{1}{6} \cdot \frac{n(n+1)}{2} \left[ n(n+1) + (2n+1) + 1 \right] = \frac{n(n+1)}{12} [n^2+n+2n+2]$.

$S_n = \frac{n(n+1)}{12} [n^2+3n+2] = \frac{n(n+1)(n+1)(n+2)}{12}$.

The sum is $\frac{n(n+1)^2(n+2)}{12}$.

Given:

The $n^{\text{th}}$ term is $a_n = n(n+1)(n+4)$.

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To Find:

The sum to $n$ terms, $S_n$.

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Solution:

First, we expand the expression for the $k^{\text{th}}$ term:

$a_k = k(k+1)(k+4) = (k^2+k)(k+4) = k^3 + 4k^2 + k^2 + 4k \ $$ = k^3 + 5k^2 + 4k$.

Now, we find the sum $S_n = \sum\limits_{k=1}^{n} a_k$.

$S_n = \sum\limits_{k=1}^{n} (k^3 + 5k^2 + 4k) = \sum\limits k^3 + 5\sum\limits k^2 + 4\sum\limits k$.

Using the standard sum formulas:

$S_n = \left(\frac{n(n+1)}{2}\right)^2 + 5\frac{n(n+1)(2n+1)}{6} + 4\frac{n(n+1)}{2}$.

Factor out the common term $\frac{n(n+1)}{12}$:

$S_n = \frac{n(n+1)}{12} [3n(n+1) + 10(2n+1) + 24]$.

$S_n = \frac{n(n+1)}{12} [3n^2+3n+20n+10+24] = \frac{n(n+1)}{12} [3n^2+23n+34]$.

Factoring the quadratic: $3n^2+23n+34 = (3n+17)(n+2)$.

$S_n = \frac{n(n+1)(n+2)(3n+17)}{12}$.

The sum is $\frac{n(n+1)(n+2)(3n+17)}{12}$.

Given:

The $n^{\text{th}}$ term is $a_n = n^2 + 2^n$.

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To Find:

The sum to $n$ terms, $S_n$.

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Solution:

The sum is $S_n = \sum\limits_{k=1}^{n} (k^2 + 2^k) = \sum\limits_{k=1}^{n} k^2 + \sum\limits_{k=1}^{n} 2^k$.

The first part is the sum of the first $n$ squares:

$\sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.

The second part is the sum of a Geometric Progression (G.P.) $2, 4, 8, \dots, 2^n$.

For this G.P., the first term is $a=2$, the common ratio is $r=2$, and there are $n$ terms.

Sum of G.P. = $\frac{a(r^n-1)}{r-1} = \frac{2(2^n-1)}{2-1} = 2(2^n-1) = 2^{n+1}-2$.

Combining the two parts:

$S_n = \frac{n(n+1)(2n+1)}{6} + 2^{n+1} - 2$.

The sum is $\frac{n(n+1)(2n+1)}{6} + 2(2^n - 1)$.

Given:

The $n^{\text{th}}$ term is $a_n = (2n-1)^2$.

---

To Find:

The sum to $n$ terms, $S_n$.

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Solution:

The $k^{\text{th}}$ term is $a_k = (2k-1)^2 = 4k^2 - 4k + 1$.

$S_n = \sum\limits_{k=1}^{n} (4k^2 - 4k + 1) = 4\sum\limits k^2 - 4\sum\limits k + \sum\limits 1$.

$S_n = 4\frac{n(n+1)(2n+1)}{6} - 4\frac{n(n+1)}{2} + n$.

$S_n = \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n$.

Factor out $n$:

$S_n = n \left[ \frac{2(n+1)(2n+1)}{3} - 2(n+1) + 1 \right]$.

$S_n = n \left[ \frac{2(2n^2+3n+1) - 6(n+1) + 3}{3} \right]$.

$S_n = \frac{n}{3} [4n^2+6n+2 - 6n-6+3] = \frac{n}{3} [4n^2-1]$.

The sum is $\frac{n(4n^2-1)}{3}$ or $\frac{n(2n-1)(2n+1)}{3}$.

Given:

Let an A.P. have first term $A$ and common difference $D$.

The $p^{\text{th}}, q^{\text{th}}, r^{\text{th}}, s^{\text{th}}$ terms of this A.P. are in G.P. Let these terms be $a_p, a_q, a_r, a_s$.

$a_p = A + (p-1)D$

$a_q = A + (q-1)D$

$a_r = A + (r-1)D$

$a_s = A + (s-1)D$

Since $a_p, a_q, a_r, a_s$ are in G.P., they have a common ratio, let's call it $R$.

$\frac{a_q}{a_p} = \frac{a_r}{a_q} = \frac{a_s}{a_r} = R$.

---

To Show:

The terms $(p-q), (q-r), (r-s)$ are in G.P.

This means we need to prove that $\frac{q-r}{p-q} = \frac{r-s}{q-r}$.

---

Proof:

From the G.P. condition, we can write:

$a_q - a_p = a_p R - a_p = a_p(R-1)$.

$a_r - a_q = a_q R - a_q = a_q(R-1)$.

$a_s - a_r = a_r R - a_r = a_r(R-1)$.

From the A.P. definition, we also have:

$a_q - a_p = (q-p)D$.

$a_r - a_q = (r-q)D$.

$a_s - a_r = (s-r)D$.

Equating the expressions for the differences:

$-(p-q)D = a_p(R-1)$

$-(q-r)D = a_q(R-1)$

$-(r-s)D = a_r(R-1)$

Now, let's find the ratio $\frac{q-r}{p-q}$:

$\frac{q-r}{p-q} = \frac{-(q-r)D}{-(p-q)D} = \frac{a_q(R-1)}{a_p(R-1)} = \frac{a_q}{a_p} = R$.

Similarly, let's find the ratio $\frac{r-s}{q-r}$:

$\frac{r-s}{q-r} = \frac{-(r-s)D}{-(q-r)D} = \frac{a_r(R-1)}{a_q(R-1)} = \frac{a_r}{a_q} = R$.

Since both ratios are equal to the common ratio $R$ of the G.P. formed by the terms, we have:

$\frac{q-r}{p-q} = \frac{r-s}{q-r}$.

This shows that $(p-q), (q-r), (r-s)$ are in G.P.

Hence, proved.

Given:

1. $a, b, c$ are in G.P., which means $b^2 = ac$.

2. $a^{1/x} = b^{1/y} = c^{1/z}$.

---

To Prove:

$x, y, z$ are in A.P., which means $2y = x+z$.

---

Proof:

Let $a^{1/x} = b^{1/y} = c^{1/z} = K$ for some constant $K$.

From this, we can express $a, b, c$ in terms of $K$:

$a = K^x$

$b = K^y$

$c = K^z$

Now, substitute these into the G.P. condition, $b^2 = ac$:

$(K^y)^2 = (K^x)(K^z)$

$K^{2y} = K^{x+z}$

Assuming $K \neq 1$ (for a non-trivial case), we can equate the exponents:

$2y = x+z$.

This is the condition for $x, y, z$ to be in an Arithmetic Progression.

Hence, proved.

Given:

The inequality $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \le 0$.

---

To Show:

$a, b, c, d$ are in G.P.

---

Proof:

We can rearrange the given expression by grouping terms with $a, b, c, d$:

$(a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bcp + c^2) + (c^2p^2 - 2cdp + d^2) \le 0$.

This can be written as a sum of squares:

$(ap - b)^2 + (bp - c)^2 + (cp - d)^2 \le 0$.

Since the square of any real number is non-negative, the sum of these squares must be greater than or equal to zero. The only way for the sum to be less than or equal to zero is if it is exactly zero. This happens only if each term is zero.

Therefore, we must have:

$(ap - b) = 0 \implies ap = b \implies p = \frac{b}{a}$.

$(bp - c) = 0 \implies bp = c \implies p = \frac{c}{b}$.

$(cp - d) = 0 \implies cp = d \implies p = \frac{d}{c}$.

From these equations, we get:

$\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$.

This is the condition for the numbers $a, b, c, d$ to be in a Geometric Progression.

Hence, proved.

Given:

1. $p, q, r$ are in G.P., so $q^2=pr$.

2. The equations $px^2+2qx+r=0$ and $dx^2+2ex+f=0$ have a common root.

---

To Show:

$\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in A.P.

---

Proof:

Consider the first equation, $px^2+2qx+r=0$. The discriminant is $\Delta = (2q)^2 - 4pr = 4q^2 - 4pr$.

Since $q^2=pr$, we have $\Delta = 4(pr) - 4pr = 0$.

A quadratic equation with a zero discriminant has one repeated real root. The root is $x = \frac{-2q}{2p} = -\frac{q}{p}$.

This must be the common root. So, we substitute $x = -\frac{q}{p}$ into the second equation:

$d\left(-\frac{q}{p}\right)^2 + 2e\left(-\frac{q}{p}\right) + f = 0$.

$d\frac{q^2}{p^2} - \frac{2eq}{p} + f = 0$.

Substitute $q^2=pr$ into this equation:

$d\frac{pr}{p^2} - \frac{2eq}{p} + f = 0 \implies \frac{dr}{p} - \frac{2eq}{p} + f = 0$.

Multiply the entire equation by $\frac{1}{r}$ (assuming $r \neq 0$):

$\frac{d}{p} - \frac{2eq}{pr} + \frac{f}{r} = 0$.

Again, substitute $pr=q^2$:

$\frac{d}{p} - \frac{2eq}{q^2} + \frac{f}{r} = 0 \implies \frac{d}{p} - \frac{2e}{q} + \frac{f}{r} = 0$.

Rearranging this gives:

$2\frac{e}{q} = \frac{d}{p} + \frac{f}{r}$.

This is the condition for $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ to be in an Arithmetic Progression.

Hence, proved.

Given:

An arithmetic progression (A.P.).

---

To Prove:

The sum of the $(m+n)$-th term and the $(m-n)$-th term is equal to twice the $m$-th term.

That is, $a_{m+n} + a_{m-n} = 2 \cdot a_m$, where $a_k$ represents the $k$-th term of the A.P.

---

Proof:

Let the first term of the A.P. be $a$ and the common difference be $d$.

The formula for the $k$-th term of an A.P. is given by:

Using this formula, we can write the expressions for the required terms:

The $(m+n)$-th term is obtained by substituting $k = m+n$ in equation (1):

The $(m-n)$-th term is obtained by substituting $k = m-n$ in equation (1). (Assume $m > n$ for the $(m-n)$-th term to be a positive indexed term. The formula holds even if $m \le n$, representing a term before the first term, which is valid in theory.):

The $m$-th term is obtained by substituting $k = m$ in equation (1):

Now, consider the sum of the $(m+n)$-th term and the $(m-n)$-th term by adding equations (2) and (3):

$a_{m+n} + a_{m-n} = [a + ((m+n)-1)d] + [a + ((m-n)-1)d]$

Remove the brackets and group terms:

$a_{m+n} + a_{m-n} = a + (m+n-1)d + a + (m-n-1)d$

$a_{m+n} + a_{m-n} = (a + a) + (m+n-1)d + (m-n-1)d$

Factor out the common difference $d$ from the terms containing $d$:

$a_{m+n} + a_{m-n} = 2a + d[(m+n-1) + (m-n-1)]$

Simplify the expression inside the square brackets:

$a_{m+n} + a_{m-n} = 2a + d[m+n-1 + m-n-1]$

$a_{m+n} + a_{m-n} = 2a + d[(m+m) + (n-n) + (-1-1)]$

$a_{m+n} + a_{m-n} = 2a + d[2m + 0 - 2]$

$a_{m+n} + a_{m-n} = 2a + d[2m - 2]$

Factor out 2 from the term containing $d$:

$a_{m+n} + a_{m-n} = 2a + 2d(m - 1)$

Now, factor out the common term 2 from the entire expression:

$a_{m+n} + a_{m-n} = 2[a + (m - 1)d]$

From equation (4), we know that $a + (m-1)d$ is the $m$-th term ($a_m$).

Substituting $a_m$ into the expression:

This result shows that the sum of the $(m+n)$-th term and the $(m-n)$-th term of an A.P. is equal to twice the $m$-th term.

---

Hence Proved.

Solution:

---

Let the three numbers in Arithmetic Progression (A.P.) be $a - d$, $a$, and $a + d$, where $a$ is the middle term and $d$ is the common difference.

---

According to the problem statement, the sum of the three numbers is 24.

$(a - d) + a + (a + d) = 24$

Simplify the equation:

$3a = 24$

Divide by 3:

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According to the problem statement, the product of the three numbers is 440.

$(a - d) \cdot a \cdot (a + d) = 440$

Rearrange the terms and use the difference of squares formula $(x-y)(x+y) = x^2 - y^2$:

$a(a^2 - d^2) = 440$

---

Now, substitute the value of $a = 8$ into the product equation:

$8(8^2 - d^2) = 440$

$8(64 - d^2) = 440$

Divide both sides by 8:

$64 - d^2 = \frac{440}{8}$

Rearrange the equation to solve for $d^2$:

$d^2 = 64 - 55$

$d^2 = 9$

Take the square root of both sides to find the value of $d$:

$d = \pm \sqrt{9}$

$d = \pm 3$

---

We have two possible values for the common difference $d$:

Case 1: $d = 3$

The three numbers are:

$a - d = 8 - 3 = 5$

$a = 8$

$a + d = 8 + 3 = 11$

The numbers are 5, 8, and 11.

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Case 2: $d = -3$

The three numbers are:

$a - d = 8 - (-3) = 8 + 3 = 11$

$a = 8$

$a + d = 8 + (-3) = 8 - 3 = 5$

The numbers are 11, 8, and 5.

---

In both cases, the set of numbers is the same {5, 8, 11}.

The sum of these numbers is $5 + 8 + 11 = 24$.

The product of these numbers is $5 \times 8 \times 11 = 40 \times 11 = 440$.

Both conditions given in the problem are satisfied.

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The numbers are 5, 8, and 11.

Given:

An Arithmetic Progression (A.P.).

$S_1$ = Sum of the first $n$ terms.

$S_2$ = Sum of the first $2n$ terms.

$S_3$ = Sum of the first $3n$ terms.

---

To Show:

$S_3 = 3(S_2 - S_1)$

---

Proof:

Let the first term of the A.P. be $a$ and the common difference be $d$.

The formula for the sum of the first $k$ terms of an A.P. is given by:

Using this formula, we can write the expressions for $S_1$, $S_2$, and $S_3$:

$S_1$ is the sum of the first $n$ terms ($k=n$):

$S_2$ is the sum of the first $2n$ terms ($k=2n$):

$S_3$ is the sum of the first $3n$ terms ($k=3n$):

Now, consider the expression $S_2 - S_1$. Subtract equation (2) from equation (3):

$S_2 - S_1 = n[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d]$

$S_2 - S_1 = \frac{n}{2} \left[ 2 \cdot (2a + (2n-1)d) - (2a + (n-1)d) \right]$

$S_2 - S_1 = \frac{n}{2} \left[ 4a + 2(2n-1)d - 2a - (n-1)d \right]$

$S_2 - S_1 = \frac{n}{2} \left[ (4a - 2a) + (4n-2)d - (n-1)d \right]$

$S_2 - S_1 = \frac{n}{2} \left[ 2a + \{(4n-2) - (n-1)\}d \right]$

$S_2 - S_1 = \frac{n}{2} \left[ 2a + (4n - 2 - n + 1)d \right]$

$S_2 - S_1 = \frac{n}{2} \left[ 2a + (3n - 1)d \right]$

Now, consider the expression $3(S_2 - S_1)$. Multiply equation (5) by 3:

$3(S_2 - S_1) = 3 \cdot \frac{n}{2} [2a + (3n - 1)d]$

Compare equation (6) with equation (4). We observe that the right-hand sides are identical.

Therefore, from equations (4) and (6), we can conclude that:

---

Hence Shown.

Solution:

---

We need to find the sum of all numbers between 200 and 400 that are divisible by 7.

These numbers form an Arithmetic Progression (A.P.) with a common difference $d = 7$.

---

Finding the first term ($a$):

The first number greater than 200 that is divisible by 7.

Divide 200 by 7: $200 = 7 \times 28 + 4$.

This means 200 is 4 more than a multiple of 7 ($7 \times 28 = 196$).

The next multiple of 7 will be $196 + 7 = 203$.

So, the first term $a = 203$.

---

Finding the last term ($l$ or $a_n$):

The last number less than 400 that is divisible by 7.

Divide 400 by 7: $400 = 7 \times 57 + 1$.

This means 400 is 1 more than a multiple of 7 ($7 \times 57 = 399$).

The last multiple of 7 before 400 is 399.

So, the last term $l = a_n = 399$.

---

Finding the number of terms ($n$):

We use the formula for the $n$-th term of an A.P.: $a_n = a + (n-1)d$.

We have $a_n = 399$, $a = 203$, and $d = 7$.

$399 = 203 + (n-1)7$

Subtract 203 from both sides:

$399 - 203 = (n-1)7$

$196 = (n-1)7$

Divide both sides by 7:

$\frac{196}{7} = n-1$

Add 1 to both sides:

$n = 28 + 1$

---

Finding the sum ($S_n$):

The sum of the first $n$ terms of an A.P. is given by the formula: $S_n = \frac{n}{2}(a + l)$.

Substitute the values $n = 29$, $a = 203$, and $l = 399$ into the formula:

$S_{29} = \frac{29}{2}(203 + 399)$

$S_{29} = \frac{29}{2}(602)$

Divide 602 by 2:

$S_{29} = 29 \times 301$

Multiply 29 by 301:

$S_{29} = 8729$

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The sum of all numbers between 200 and 400 which are divisible by 7 is 8729.

Solution:

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We want to find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Using the Principle of Inclusion-Exclusion, the required sum is:

Sum (divisible by 2 or 5) = Sum (divisible by 2) + Sum (divisible by 5) - Sum (divisible by both 2 and 5)

Numbers divisible by both 2 and 5 are those divisible by the Least Common Multiple (LCM) of 2 and 5, which is 10.

So, the required sum is: $S_{\text{by 2 or 5}} = S_{\text{by 2}} + S_{\text{by 5}} - S_{\text{by 10}}$.

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1. Sum of integers from 1 to 100 divisible by 2:

These are 2, 4, 6, ..., 100.

This is an Arithmetic Progression (A.P.) with first term $a = 2$, common difference $d = 2$, and last term $l = 100$.

To find the number of terms ($n_2$), we use the formula $l = a + (n-1)d$:

$100 = 2 + (n_2-1)2$

$98 = (n_2-1)2$

$\frac{98}{2} = n_2-1$

$49 = n_2-1$

The sum ($S_{\text{by 2}}$) is given by $S_n = \frac{n}{2}(a + l)$:

$S_{\text{by 2}} = \frac{50}{2}(2 + 100)$

$S_{\text{by 2}} = 25 \times 102$

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2. Sum of integers from 1 to 100 divisible by 5:

These are 5, 10, 15, ..., 100.

This is an A.P. with first term $a = 5$, common difference $d = 5$, and last term $l = 100$.

To find the number of terms ($n_5$), we use the formula $l = a + (n-1)d$:

$100 = 5 + (n_5-1)5$

$95 = (n_5-1)5$

$\frac{95}{5} = n_5-1$

$19 = n_5-1$

The sum ($S_{\text{by 5}}$) is given by $S_n = \frac{n}{2}(a + l)$:

$S_{\text{by 5}} = \frac{20}{2}(5 + 100)$

$S_{\text{by 5}} = 10 \times 105$

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3. Sum of integers from 1 to 100 divisible by 10:

These are 10, 20, 30, ..., 100.

This is an A.P. with first term $a = 10$, common difference $d = 10$, and last term $l = 100$.

To find the number of terms ($n_{10}$), we use the formula $l = a + (n-1)d$:

$100 = 10 + (n_{10}-1)10$

$90 = (n_{10}-1)10$

$\frac{90}{10} = n_{10}-1$

$9 = n_{10}-1$

The sum ($S_{\text{by 10}}$) is given by $S_n = \frac{n}{2}(a + l)$:

$S_{\text{by 10}} = \frac{10}{2}(10 + 100)$

$S_{\text{by 10}} = 5 \times 110$

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4. Calculate the required sum:

Using the Inclusion-Exclusion Principle and the sums calculated in (i), (ii), and (iii):

$S_{\text{by 2 or 5}} = S_{\text{by 2}} + S_{\text{by 5}} - S_{\text{by 10}}$

$S_{\text{by 2 or 5}} = 2550 + 1050 - 550$

$S_{\text{by 2 or 5}} = 3600 - 550$

$S_{\text{by 2 or 5}} = 3050$

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The sum of integers from 1 to 100 that are divisible by 2 or 5 is 3050.

Solution:

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We are looking for two-digit numbers that leave a remainder of 1 when divided by 4.

A number $x$ leaves a remainder of 1 when divided by 4 can be written in the form $x = 4k + 1$, where $k$ is an integer.

The two-digit numbers range from 10 to 99, inclusive.

---

Finding the first term:

We need to find the smallest integer $k$ such that $4k + 1 \ge 10$.

$4k \ge 10 - 1$

$4k \ge 9$

$k \ge \frac{9}{4}$

$k \ge 2.25$

Since $k$ must be an integer, the smallest possible value for $k$ is 3.

The first number in the sequence is $4(3) + 1 = 12 + 1 = 13$.

So, the first term $a = 13$.

---

Finding the last term:

We need to find the largest integer $k$ such that $4k + 1 \le 99$.

$4k \le 99 - 1$

$4k \le 98$

$k \le \frac{98}{4}$

$k \le 24.5$

Since $k$ must be an integer, the largest possible value for $k$ is 24.

The last number in the sequence is $4(24) + 1 = 96 + 1 = 97$.

So, the last term $l = 97$.

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The sequence of numbers is 13, 17, 21, ..., 97.

This is an Arithmetic Progression (A.P.) with the first term $a = 13$ and the common difference $d = 4$ (since each number is of the form $4k+1$, the difference between consecutive terms is $(4(k+1)+1) - (4k+1) = 4$).

---

Finding the number of terms ($n$):

We use the formula for the $n$-th term of an A.P.: $l = a + (n-1)d$.

Substitute the values $l = 97$, $a = 13$, and $d = 4$:

$97 = 13 + (n-1)4$

Subtract 13 from both sides:

$97 - 13 = (n-1)4$

$84 = (n-1)4$

Divide both sides by 4:

$\frac{84}{4} = n-1$

Add 1 to both sides:

$n = 21 + 1$

---

Finding the sum ($S_n$):

The sum of the first $n$ terms of an A.P. is given by the formula: $S_n = \frac{n}{2}(a + l)$.

Substitute the values $n = 22$, $a = 13$, and $l = 97$ into the formula:

$S_{22} = \frac{22}{2}(13 + 97)$

$S_{22} = 11(110)$

$S_{22} = 1210$

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The sum of all two-digit numbers which when divided by 4, yields 1 as remainder is 1210.

Solution:

---

The given functional equation is $f(x + y) = f(x) f(y)$ for all $x, y \in \mathbb{N}$.

This type of functional equation suggests that the function $f(x)$ is of the form $f(x) = a^x$ for some base $a$.

We are given that $f(1) = 3$. Substitute $x=1$ into equation (1):

$f(1) = a^1 = a$

Since $f(1) = 3$, we have:

Therefore, the function is:

---

We are also given that $\sum\limits_{x=1}^{n} f(x) = 120$.

Substitute $f(x) = 3^x$ from equation (2) into the summation:

The summation on the left side of equation (3) is $3^1 + 3^2 + 3^3 + \dots + 3^n$.

This is a geometric progression (G.P.) with:

First term, $A = 3^1 = 3$.

Common ratio, $R = \frac{3^2}{3^1} = 3$.

Number of terms = $n$.

The sum of the first $n$ terms of a G.P. with first term $A$ and common ratio $R$ (where $R \neq 1$) is given by the formula:

Substitute the values $S_n = 120$, $A = 3$, and $R = 3$ into equation (4):

$120 = 3 \frac{3^n - 1}{3 - 1}$

Now, we need to solve equation (5) for $n$.

Multiply both sides by 2:

$120 \times 2 = 3 (3^n - 1)$

$240 = 3 (3^n - 1)$

Divide both sides by 3:

$\frac{240}{3} = 3^n - 1$

Add 1 to both sides of equation (6):

$80 + 1 = 3^n$

To solve for $n$, express 81 as a power of 3:

$81 = 3 \times 27 = 3 \times 3^3 = 3^4$

Substitute this into equation (7):

Since the bases are equal, the exponents must be equal:

$n = 4$

---

The value of n is 4.

Solution:

---

Given:

Sum of terms of a G.P., $S_n = 315$.

First term, $a = 5$.

Common ratio, $r = 2$.

---

To Find:

The number of terms, $n$.

The last term, $a_n$.

---

Solution:

The sum of the first $n$ terms of a Geometric Progression (G.P.) with first term $a$ and common ratio $r$ is given by the formula:

Substitute the given values $S_n = 315$, $a = 5$, and $r = 2$ into equation (1):

$315 = 5 \frac{2^n - 1}{2 - 1}$

$315 = 5 \frac{2^n - 1}{1}$

$315 = 5 (2^n - 1)$

Divide both sides by 5:

$\frac{315}{5} = 2^n - 1$

Add 1 to both sides of equation (2):

$63 + 1 = 2^n$

Express 64 as a power of 2:

$64 = 2 \times 32 = 2 \times 2 \times 16 = 2 \times 2 \times 2 \times 8 = 2^4 \times 4 = 2^4 \times 2^2 = 2^6$

Substitute $64 = 2^6$ into equation (3):

Since the bases are equal in equation (4), the exponents must be equal:

---

Now, we need to find the last term, which is the $n$-th term ($a_n$).

The formula for the $n$-th term of a G.P. is:

Substitute the values $a = 5$, $r = 2$, and $n = 6$ into equation (5) to find the 6th term ($a_6$):

$a_6 = 5 \times 2^{6-1}$

$a_6 = 5 \times 2^5$

Calculate $2^5$:

$2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$

Substitute $2^5 = 32$ back into the equation for $a_6$:

Calculate the product:

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The number of terms is 6 and the last term is 160.

Solution:

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Given:

The first term of a Geometric Progression (G.P.), $a = 1$.

The sum of the third term ($a_3$) and the fifth term ($a_5$) is 90.

$a_3 + a_5 = 90$

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To Find:

The common ratio, $r$, of the G.P.

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The formula for the $n$-th term of a Geometric Progression (G.P.) with first term $a$ and common ratio $r$ is given by:

Using this formula, we can express the third term ($a_3$) and the fifth term ($a_5$) in terms of $a$ and $r$:

For the third term, substitute $n=3$ into equation (1):

For the fifth term, substitute $n=5$ into equation (1):

We are given that the first term $a = 1$. Substitute $a=1$ into equations (2) and (3):

$a_3 = (1)r^2 = r^2$

$a_5 = (1)r^4 = r^4$

We are also given that the sum of the third and fifth terms is 90:

Substitute the expressions for $a_3$ and $a_5$ into this equation:

Rearrange equation (4) into a standard quadratic form in terms of $r^2$:

$r^4 + r^2 - 90 = 0$

Let $x = r^2$. The equation becomes a quadratic equation in $x$:

We can solve this quadratic equation by factoring. We look for two numbers that multiply to -90 and add up to 1. These numbers are 10 and -9.

So, equation (5) can be factored as:

$(x + 10)(x - 9) = 0$

This gives two possible values for $x$:

$x + 10 = 0$ or $x - 9 = 0$

$x = -10$ $x = 9$

Now, substitute back $x = r^2$:

$r^2 = -10$ or $r^2 = 9$

For real values of $r$, $r^2$ must be non-negative. Thus, $r^2 = -10$ has no real solutions for $r$.

Consider the second possibility:

$r^2 = 9$

Taking the square root of both sides:

$r = \pm \sqrt{9}$

$r = \pm 3$

So, the possible real values for the common ratio are 3 and -3.

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The common ratio of the G.P. can be 3 or -3.

Given:

Let the three numbers in G.P. be $a, ar, ar^2$.

1. Sum of the numbers: $a + ar + ar^2 = 56$.

2. The numbers $a-1, ar-7, ar^2-21$ are in A.P.

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To Find:

The three numbers.

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Solution:

From the first condition: $a(1+r+r^2) = 56$. ...(i)

From the second condition, if three numbers are in A.P., then twice the middle term equals the sum of the other two.

$2(ar-7) = (a-1) + (ar^2-21)$.

$2ar - 14 = a + ar^2 - 22$.

$8 = a + ar^2 - 2ar = a(1+r^2-2r) = a(r-1)^2$. ...(ii)

From (i), $a = \frac{56}{1+r+r^2}$. Substitute this into (ii):

$8 = \frac{56}{1+r+r^2}(r-1)^2 \implies \frac{8}{56} = \frac{(r-1)^2}{1+r+r^2}$.

$\frac{1}{7} = \frac{r^2-2r+1}{r^2+r+1} \implies r^2+r+1 = 7(r^2-2r+1)$.

$r^2+r+1 = 7r^2-14r+7 \implies 6r^2 - 15r + 6 = 0$.

Dividing by 3: $2r^2-5r+2=0$.

Factoring: $(2r-1)(r-2)=0$. So, $r=2$ or $r=1/2$.

Case 1: $r=2$

From (ii), $a(2-1)^2=8 \implies a=8$. The numbers are $8, 8(2), 8(2^2) \implies 8, 16, 32$.

Case 2: $r=1/2$

From (ii), $a(\frac{1}{2}-1)^2=8 \implies a(-\frac{1}{2})^2=8 \implies \frac{a}{4}=8 \implies a=32$.

The numbers are $32, 32(\frac{1}{2}), 32(\frac{1}{2})^2 \implies 32, 16, 8$.

In both cases, the set of numbers is the same.

The numbers are 8, 16, 32.

Given:

Let the G.P. have an even number of terms, say $2n$.

Sum of all terms ($S_{2n}$) = 5 $\times$ (Sum of terms at odd places).

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To Find:

The common ratio, $r$.

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Solution:

Let the first term be $a$ and common ratio be $r$.

The terms are $a, ar, ar^2, \dots, ar^{2n-1}$.

The sum of all terms is $S_{2n} = \frac{a(r^{2n}-1)}{r-1}$.

The terms at odd places are $a_1, a_3, a_5, \dots, a_{2n-1}$, which are $a, ar^2, ar^4, \dots$.

This is a G.P. with first term $a$, common ratio $r^2$, and $n$ terms.

Sum of terms at odd places = $\frac{a((r^2)^n-1)}{r^2-1} = \frac{a(r^{2n}-1)}{r^2-1}$.

Given the condition: $\frac{a(r^{2n}-1)}{r-1} = 5 \cdot \frac{a(r^{2n}-1)}{r^2-1}$.

Assuming a non-trivial G.P. where $a \neq 0$ and $r^{2n} \neq 1$, we can cancel $\frac{a(r^{2n}-1)}{1}$ from both sides.

$\frac{1}{r-1} = \frac{5}{r^2-1} = \frac{5}{(r-1)(r+1)}$.

Since $r \neq 1$, we can cancel $(r-1)$:

$1 = \frac{5}{r+1} \implies r+1=5 \implies r=4$.

The common ratio is 4.

Given:

An Arithmetic Progression (A.P.) with:

First term, $a = 11$.

Sum of the first four terms, $S_4 = 56$.

Sum of the last four terms = 112.

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To Find:

The total number of terms, $n$, in the A.P.

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Solution:

Let the common difference of the A.P. be $d$.

The first four terms of the A.P. are $a, (a+d), (a+2d), (a+3d)$.

It is given that their sum is 56.

$a + (a+d) + (a+2d) + (a+3d) = 56$

$4a + 6d = 56$

Substitute the given value of the first term, $a=11$:

$4(11) + 6d = 56$

$44 + 6d = 56$

$6d = 56 - 44$

$6d = 12$

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Let the total number of terms in the A.P. be $n$.

The last four terms are the $n^{th}$, $(n-1)^{th}$, $(n-2)^{th}$, and $(n-3)^{th}$ terms.

The $k^{th}$ term of an A.P. is given by $a_k = a + (k-1)d$.

Last term, $a_n = a + (n-1)d = 11 + (n-1)2$

Second last term, $a_{n-1} = a + (n-2)d = 11 + (n-2)2$

Third last term, $a_{n-2} = a + (n-3)d = 11 + (n-3)2$

Fourth last term, $a_{n-3} = a + (n-4)d = 11 + (n-4)2$

The sum of these last four terms is given as 112.

$[11 + (n-1)2] + [11 + (n-2)2] + [11 + (n-3)2] \ $$ + [11 + (n-4)2] = 112$

$4(11) + 2[(n-1) + (n-2) + (n-3) + (n-4)] = 112$

$44 + 2[4n - 10] = 112$

$44 + 8n - 20 = 112$

$24 + 8n = 112$

$8n = 112 - 24$

$8n = 88$

$n = \frac{88}{8} = 11$

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Thus, the total number of terms in the A.P. is 11.

Given:

$\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}$, where $x \neq 0$.

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To Show:

$a, b, c,$ and $d$ are in Geometric Progression (G.P.).

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Proof:

For $a, b, c, d$ to be in G.P., we must show that their common ratio is constant, i.e., $\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$. This is equivalent to proving that $b^2 = ac$ and $c^2 = bd$.

First, consider the equality:

Applying cross-multiplication:

$(a + bx)(b - cx) = (b + cx)(a - bx)$

$ab - acx + b^2x - bcx^2 = ab - b^2x + acx - bcx^2$

Cancelling $ab$ and $-bcx^2$ from both sides:

$-acx + b^2x = -b^2x + acx$

$2b^2x = 2acx$

Since it is given that $x \neq 0$, we can divide by $2x$:

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Next, consider the equality:

Applying cross-multiplication:

$(b + cx)(c - dx) = (c + dx)(b - cx)$

$bc - bdx + c^2x - cdx^2 = bc - c^2x + bdx - cdx^2$

Cancelling $bc$ and $-cdx^2$ from both sides:

$-bdx + c^2x = -c^2x + bdx$

$2c^2x = 2bdx$

Since $x \neq 0$, we can divide by $2x$:

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From (i), $b^2 = ac \implies \frac{b}{a} = \frac{c}{b}$.

From (ii), $c^2 = bd \implies \frac{c}{b} = \frac{d}{c}$.

Combining these results gives $\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$.

This shows that the ratio between consecutive terms is constant. Therefore, $a, b, c,$ and $d$ are in G.P.

Hence Proved.

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Alternate Solution (Using Componendo and Dividendo):

The rule of Componendo and Dividendo states that if $\frac{p}{q} = \frac{r}{s}$, then $\frac{p+q}{p-q} = \frac{r+s}{r-s}$.

Applying this rule to $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx}$:

$\frac{(a + bx) + (a - bx)}{(a + bx) - (a - bx)} = \frac{(b + cx) + (b - cx)}{(b + cx) - (b - cx)}$

$\frac{2a}{2bx} = \frac{2b}{2cx}$

Since $x \neq 0$, we can cancel $2x$ from the denominators: $\frac{a}{b} = \frac{b}{c} \implies b^2 = ac$.

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Similarly, applying the rule to $\frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}$:

$\frac{(b + cx) + (b - cx)}{(b + cx) - (b - cx)} = \frac{(c + dx) + (c - dx)}{(c + dx) - (c - dx)}$

$\frac{2b}{2cx} = \frac{2c}{2dx}$

Since $x \neq 0$, we can cancel $2x$ from the denominators: $\frac{b}{c} = \frac{c}{d} \implies c^2 = bd$.

From $b^2 = ac$ and $c^2=bd$, we conclude that $a,b,c,d$ are in G.P.

Given:

A Geometric Progression (G.P.) of $n$ terms.

$S$ = Sum of the $n$ terms.

$P$ = Product of the $n$ terms.

$R$ = Sum of the reciprocals of the $n$ terms.

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To Prove:

$P^2 R^n = S^n$

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Proof:

Let the G.P. be $a, ar, ar^2, \dots, ar^{n-1}$, where $a$ is the first term and $r$ is the common ratio. We assume $a \neq 0, r \neq 0, r \neq 1$.

Step 1: Express S, P, and R in terms of a and r.

The sum of the terms is:

The product of the terms is:

$P = a \cdot ar \cdot ar^2 \cdots ar^{n-1} = a^n r^{0+1+2+\dots+(n-1)}$

The sum of the reciprocals is:

$R = \frac{1}{a} + \frac{1}{ar} + \dots + \frac{1}{ar^{n-1}}$

This is a G.P. with first term $\frac{1}{a}$ and common ratio $\frac{1}{r}$.

$R = \frac{1}{a} \left(\frac{(1/r)^n - 1}{1/r - 1}\right) = \frac{1}{a} \left(\frac{\frac{1-r^n}{r^n}}{\frac{1-r}{r}}\right) = \frac{1}{a} \frac{1-r^n}{r^n} \frac{r}{1-r}$

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Step 2: Establish a relationship between S and R.

From (i), we can write $\frac{r^n-1}{r-1} = \frac{S}{a}$.

Substitute this into our expression for R from (iii):

$R = \frac{1}{a} \left(\frac{S}{a}\right) \frac{1}{r^{n-1}} = \frac{S}{a^2 r^{n-1}}$

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Step 3: Evaluate $P^2 R^n$ using the expressions above.

From (ii), we square P:

$P^2 = \left( a^n r^{\frac{n(n-1)}{2}} \right)^2 = a^{2n} r^{n(n-1)}$

From our relationship between S and R, we raise R to the power of n:

$R^n = \left(\frac{S}{a^2 r^{n-1}}\right)^n = \frac{S^n}{(a^2)^n (r^{n-1})^n} = \frac{S^n}{a^{2n} r^{n(n-1)}}$

Now, we compute the product $P^2 R^n$:

$P^2 R^n = \left( a^{2n} r^{n(n-1)} \right) \cdot \left( \frac{S^n}{a^{2n} r^{n(n-1)}} \right)$

The terms $a^{2n}$ and $r^{n(n-1)}$ cancel out:

$P^2 R^n = S^n$

The proof holds. For the case $r=1$, $S=na$, $P=a^n$, $R=n/a$. Then $P^2 R^n = (a^n)^2(n/a)^n = a^{2n} n^n / a^n = n^n a^n = (na)^n = S^n$. The identity is true for all cases.

Hence Proved.

Given:

In an Arithmetic Progression (A.P.):

The $p^{th}$ term is $a$.

The $q^{th}$ term is $b$.

The $r^{th}$ term is $c$.

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To Show:

$(q – r)a + (r – p)b + (p – q)c = 0$

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Proof:

Let the first term of the A.P. be $A$ and the common difference be $D$.

According to the definition of an A.P., we can write the given conditions as a system of equations:

To simplify, we can eliminate $A$ by subtracting the equations from one another.

Subtracting (ii) from (i):

$a - b = [A + (p-1)D] - [A + (q-1)D]$

$a - b = (p-1 - (q-1))D = (p-q)D$

Subtracting (iii) from (ii):

$b - c = [A + (q-1)D] - [A + (r-1)D]$

$b - c = (q-1 - (r-1))D = (q-r)D$

Subtracting (i) from (iii):

$c - a = [A + (r-1)D] - [A + (p-1)D]$

$c - a = (r-1 - (p-1))D = (r-p)D$

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Now, let's consider the Left Hand Side (LHS) of the expression we need to prove:

LHS = $(q – r)a + (r – p)b + (p – q)c$

Substitute the expressions for $(q-r)$, $(r-p)$, and $(p-q)$ from equations (v), (vi), and (iv) respectively (assuming $D \neq 0$):

LHS = $\left(\frac{b-c}{D}\right)a + \left(\frac{c-a}{D}\right)b + \left(\frac{a-b}{D}\right)c$

Factor out $\frac{1}{D}$:

LHS = $\frac{1}{D} [a(b-c) + b(c-a) + c(a-b)]

LHS = $\frac{1}{D} [ab - ac + bc - ab + ac - bc]$

LHS = $\frac{1}{D} [0] = 0$

LHS = RHS

Note: If $D=0$, then $a=b=c$. The expression becomes $(q-r)a + (r-p)a + (p-q)a = a(q-r+r-p+p-q) = a(0) = 0$. So the identity holds in all cases.

Hence Proved.

Given:

The terms $a\left( \frac{1}{b}+\frac{1}{c} \right)$, $b\left( \frac{1}{c}+\frac{1}{a} \right)$, and $c\left( \frac{1}{a}+\frac{1}{b} \right)$ are in A.P.

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To Prove:

$a$, $b$, $c$ are in A.P.

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Proof:

Let the given terms of the A.P. be $T_1, T_2, T_3$.

$T_1 = a\left( \frac{1}{b}+\frac{1}{c} \right) = \frac{a}{b} + \frac{a}{c}$

$T_2 = b\left( \frac{1}{c}+\frac{1}{a} \right) = \frac{b}{c} + \frac{b}{a}$

$T_3 = c\left( \frac{1}{a}+\frac{1}{b} \right) = \frac{c}{a} + \frac{c}{b}$

A property of an A.P. is that if a constant is added to each term, the resulting sequence is also an A.P.

Let's add 1 to each term:

$T_1 + 1 = \frac{a}{b} + \frac{a}{c} + 1 = \frac{ac+ab+bc}{bc}$

$T_2 + 1 = \frac{b}{c} + \frac{b}{a} + 1 = \frac{ab+bc+ac}{ac}$

$T_3 + 1 = \frac{c}{a} + \frac{c}{b} + 1 = \frac{bc+ac+ab}{ab}$

So, the terms $\frac{ab+bc+ca}{bc}$, $\frac{ab+bc+ca}{ac}$, and $\frac{ab+bc+ca}{ab}$ are in A.P.

Another property of an A.P. is that if each term is multiplied by a non-zero constant, the resulting sequence is also an A.P.

Let's multiply each term by $\frac{1}{ab+bc+ca}$ (assuming $ab+bc+ca \neq 0$):

The new terms are $\frac{1}{bc}$, $\frac{1}{ac}$, $\frac{1}{ab}$. These are also in A.P.

Now, let's multiply these terms by the non-zero constant $abc$ (assuming $a,b,c \neq 0$):

$\frac{abc}{bc}$, $\frac{abc}{ac}$, $\frac{abc}{ab}$ are in A.P.

Simplifying these terms gives:

$a, b, c$

Thus, $a, b, c$ are in A.P.

Hence Proved.

Given:

The terms $a, b, c, d$ are in Geometric Progression (G.P.).

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To Prove:

The terms $(a^n + b^n)$, $(b^n + c^n)$, and $(c^n + d^n)$ are in G.P.

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Proof:

Since $a, b, c, d$ are in G.P., let the common ratio be $r$. We can express the terms as:

$b = ar$

$c = ar^2$

$d = ar^3$

To prove that $(a^n + b^n)$, $(b^n + c^n)$, and $(c^n + d^n)$ are in G.P., we need to show that the ratio of consecutive terms is constant.

Let's calculate the ratio of the second term to the first term:

$\frac{b^n + c^n}{a^n + b^n} = \frac{(ar)^n + (ar^2)^n}{a^n + (ar)^n}$

$\frac{a^n r^n + a^n r^{2n}}{a^n + a^n r^n} = \frac{a^n r^n (1 + r^n)}{a^n (1 + r^n)}$

Assuming $a \neq 0$ and $1+r^n \neq 0$, we can cancel the common factors:

Now, let's calculate the ratio of the third term to the second term:

$\frac{c^n + d^n}{b^n + c^n} = \frac{(ar^2)^n + (ar^3)^n}{(ar)^n + (ar^2)^n}$

$\frac{a^n r^{2n} + a^n r^{3n}}{a^n r^n + a^n r^{2n}} = \frac{a^n r^{2n} (1 + r^n)}{a^n r^n (1 + r^n)}$

Cancelling the common factors:

From (i) and (ii), we see that the ratio of consecutive terms is constant and equal to $r^n$.

$\frac{b^n + c^n}{a^n + b^n} = \frac{c^n + d^n}{b^n + c^n} = r^n$

Therefore, the terms $(a^n + b^n)$, $(b^n + c^n)$, and $(c^n + d^n)$ form a G.P.

Hence Proved.

Given:

$a, b$ are the roots of $x^2 - 3x + p = 0$.

$c, d$ are the roots of $x^2 - 12x + q = 0$.

$a, b, c, d$ form a G.P.

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To Prove:

$(q + p) : (q - p) = 17:15$.

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Proof:

From the properties of quadratic equations (Vieta's formulas), for the equation $x^2 - 3x + p = 0$:

For the equation $x^2 - 12x + q = 0$:

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Since $a, b, c, d$ are in a G.P., let the first term be $a$ and the common ratio be $r$. Then:

$b = ar, \quad c = ar^2, \quad d = ar^3$

Substitute these into the sum equations (i) and (iii):

From (i): $a + ar = 3 \implies a(1+r) = 3$

From (iii): $ar^2 + ar^3 = 12 \implies ar^2(1+r) = 12$

Now, divide the second resulting equation by the first:

$\frac{ar^2(1+r)}{a(1+r)} = \frac{12}{3}$

$r^2 = 4$

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Now, we express $p$ and $q$ in terms of $a$ and $r$ using the product equations (ii) and (iv):

From (ii): $p = ab = a(ar) = a^2r$

From (iv): $q = cd = (ar^2)(ar^3) = a^2r^5$

We need to find the ratio $\frac{q+p}{q-p}$.

$\frac{q+p}{q-p} = \frac{a^2r^5 + a^2r}{a^2r^5 - a^2r}$

Factor out the common term $a^2r$ from the numerator and denominator:

$\frac{q+p}{q-p} = \frac{a^2r(r^4 + 1)}{a^2r(r^4 - 1)} = \frac{r^4 + 1}{r^4 - 1}$

We know that $r^2 = 4$. Therefore, $r^4 = (r^2)^2 = 4^2 = 16$.

Substitute this value back into the ratio:

$\frac{q+p}{q-p} = \frac{16 + 1}{16 - 1} = \frac{17}{15}$

This means $(q + p) : (q - p) = 17:15$.

Hence Proved.

Given:

$a$ and $b$ are two positive numbers.

Ratio of A.M. to G.M. is $m:n$.

A.M. $= \frac{a+b}{2}$

G.M. $= \sqrt{ab}$

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To Show:

$a : b = \left( m+\sqrt{m^{2}-n^{2}} \right) : \left( m-\sqrt{m^{2}-n^{2}} \right)$

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Proof:

According to the given information, the ratio of the A.M. and G.M. of $a$ and $b$ is $m:n$.

Since $a$ and $b$ are positive and $a \neq b$ (otherwise A.M. = G.M., $m=n$, and the required ratio would involve $\sqrt{0}$), A.M. > G.M., which implies $m > n$. Thus, $m^2 - n^2 > 0$.

From equation (i), squaring both sides gives:

We know that $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$. Adding $4ab$ to $(a-b)^2$ gives $(a-b)^2 + 4ab = a^2 - 2ab + b^2 + 4ab = a^2 + 2ab + b^2 = (a+b)^2$. So, $(a+b)^2 - 4ab = (a-b)^2$.

From $\frac{(a+b)^2}{4ab} = \frac{m^2}{n^2}$, subtract 1 from both sides:

$\frac{(a+b)^2}{4ab} - 1 = \frac{m^2}{n^2} - 1$

$\frac{(a+b)^2 - 4ab}{4ab} = \frac{m^2 - n^2}{n^2}$

Substitute $(a+b)^2 - 4ab = (a-b)^2$:

Taking the square root of both sides (since $a,b$ are positive, $\sqrt{ab}$ is positive, and assuming $a>b$, $a-b$ is positive, so we take the positive root on the RHS):

Now, divide equation (i) by equation (ii):

Now, apply Componendo and Dividendo to $\frac{a+b}{a-b} = \frac{m}{\sqrt{m^2 - n^2}}$:

Simplifying the left-hand side:

$\frac{a+b+a-b}{a+b-a+b} = \frac{2a}{2b} = \frac{a}{b}$

So, we get:

This can be written in ratio form as:

$a : b = \left( m+\sqrt{m^{2}-n^{2}} \right) : \left( m-\sqrt{m^{2}-n^{2}} \right)$

Hence, proved.

Given:

1. $a, b, c$ are in A.P.

2. $b, c, d$ are in G.P.

3. $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in A.P.

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To Prove:

$a, c, e$ are in G.P.

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Proof:

Since $a, b, c$ are in A.P., we have:

Since $b, c, d$ are in G.P., we have:

Since $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in A.P., we have:

Taking the reciprocal of both sides (assuming $d, c, e, c+e$ are non-zero):

Multiplying by 2:

Substitute the value of $b$ from (i) into (ii): $b = \frac{a+c}{2}$.

Now substitute the value of $d$ from (iii) into (iv):

Simplify the right-hand side:

Assuming $c \neq 0$, divide both sides by $c$:

Multiply both sides by $(c+e)$ (assuming $c+e \neq 0$):

Expand both sides:

Subtract $ce$ from both sides:

The condition $c^2 = ae$ implies that $a, c, e$ are in G.P.

Therefore, $a, c, e$ are in G.P.

(i) 5 + 55 + 555 + … up to n terms

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To Find:

The sum of the series $5 + 55 + 555 + \dots$ up to $n$ terms.

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Solution:

Let $S_n$ be the sum of the first $n$ terms of the series.

$S_n = 5 + 55 + 555 + \dots$ to $n$ terms

Take 5 common from all the terms:

$S_n = 5(1 + 11 + 111 + \dots$ to $n$ terms)

Multiply and divide by 9:

$S_n = \frac{5}{9}(9 + 99 + 999 + \dots$ to $n$ terms)

Rewrite the terms inside the bracket as a difference from powers of 10:

$S_n = \frac{5}{9}((10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots$ to $n$ terms)

Separate the terms into two groups:

$S_n = \frac{5}{9}[(10 + 10^2 + 10^3 + \dots$ to $n$ terms) - $(1 + 1 + 1 + \dots$ to $n$ terms)]

The first group is a Geometric Progression (G.P.) with first term $a=10$, common ratio $r=10$, and $n$ terms. The sum of this G.P. is $a\frac{r^n-1}{r-1} = 10\frac{10^n-1}{10-1} = \frac{10}{9}(10^n - 1)$.

The second group is the sum of $n$ ones, which is $n$.

Substitute these sums back into the expression for $S_n$:

$S_n = \frac{5}{9} \left[ \frac{10}{9}(10^n - 1) - n \right]$

$S_n = \frac{5}{81} [10(10^n - 1) - 9n]$

$S_n = \frac{5}{81} [10^{n+1} - 10 - 9n]$

The sum of the series is $\frac{5}{81}(10^{n+1} - 9n - 10)$.

---

(ii) 0.6 + 0.66 + 0.666 + … up to n terms

---

To Find:

The sum of the series $0.6 + 0.66 + 0.666 + \dots$ up to $n$ terms.

---

Solution:

Let $S_n$ be the sum of the first $n$ terms of the series.

$S_n = 0.6 + 0.66 + 0.666 + \dots$ to $n$ terms

Take 6 common from all the terms:

$S_n = 6(0.1 + 0.11 + 0.111 + \dots$ to $n$ terms)

Multiply and divide by 9:

$S_n = \frac{6}{9}(0.9 + 0.99 + 0.999 + \dots$ to $n$ terms)

$S_n = \frac{2}{3}((1 - 0.1) + (1 - 0.01) + (1 - 0.001) + \dots$ to $n$ terms)

Separate the terms into two groups:

$S_n = \frac{2}{3}[(1 + 1 + \dots$ to $n$ terms) - $(0.1 + 0.01 + 0.001 + \dots$ to $n$ terms)]

The first group is the sum of $n$ ones, which is $n$.

The second group is a G.P. with first term $a=0.1=\frac{1}{10}$, common ratio $r=\frac{1}{10}$, and $n$ terms. The sum of this G.P. is $a\frac{1-r^n}{1-r} = \frac{1}{10} \frac{1-(1/10)^n}{1-1/10} = \frac{1/10}{9/10}(1 - 10^{-n}) = \frac{1}{9}(1 - 10^{-n})$.

Substitute these sums back into the expression for $S_n$:

$S_n = \frac{2}{3} \left[ n - \frac{1}{9}(1 - 10^{-n}) \right]$

The sum of the series is $\frac{2}{3}n - \frac{2}{27}(1 - 10^{-n})$.

Given:

The series $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots$

---

To Find:

The 20th term of the series.

---

Solution:

Let the $n^{th}$ term of the series be denoted by $a_n$.

First, we observe the pattern of the terms in the series.

$a_1 = 2 \times 4$

$a_2 = 4 \times 6$

$a_3 = 6 \times 8$

The $n^{th}$ term $a_n$ is a product of two numbers.

The first factor of each term forms a sequence: $2, 4, 6, \dots$. This is an A.P. with first term 2 and common difference 2. The $n^{th}$ term of this A.P. is $2 + (n-1)2 = 2n$.

The second factor of each term forms a sequence: $4, 6, 8, \dots$. This is an A.P. with first term 4 and common difference 2. The $n^{th}$ term of this A.P. is $4 + (n-1)2 = 2n+2$.

Therefore, the general expression for the $n^{th}$ term of the given series is:

$a_n = (n^{th} \text{ term of first sequence}) \times (n^{th} \text{ term of second sequence})$

$a_n = (2n)(2n+2)$

To find the 20th term, we substitute $n = 20$ into the general formula:

$a_{20} = (2 \times 20)(2 \times 20 + 2)$

$a_{20} = (40)(40 + 2)$

$a_{20} = 40 \times 42$

$a_{20} = 1680$

---

The 20th term of the series is 1680.

Given:

The series is $3, 7, 13, 21, 31, \dots$

---

To Find:

The sum of the first $n$ terms of the series, $S_n$.

---

Solution:

Step 1: Find the general term ($a_n$) of the series.

Let the given series be $a_1, a_2, a_3, \dots$.

We check the differences between consecutive terms:

$a_2 - a_1 = 7 - 3 = 4$

$a_3 - a_2 = 13 - 7 = 6$

$a_4 - a_3 = 21 - 13 = 8$

$a_5 - a_4 = 31 - 21 = 10$

The first differences $4, 6, 8, 10, \dots$ are in an A.P. This means the general term of the series is a quadratic in $n$. Let $a_n = An^2 + Bn + C$.

For $n=1: A(1)^2+B(1)+C = 3 \implies A+B+C=3$

For $n=2: A(2)^2+B(2)+C = 7 \implies 4A+2B+C=7$

For $n=3: A(3)^2+B(3)+C = 13 \implies 9A+3B+C=13$

Subtracting consecutive equations:

$(4A+2B+C) - (A+B+C) = 7-3 \implies 3A+B=4$

$(9A+3B+C) - (4A+2B+C) = 13-7 \implies 5A+B=6$

Subtracting these new equations:

$(5A+B) - (3A+B) = 6-4 \implies 2A=2 \implies A=1$

Substituting $A=1$ into $3A+B=4$ gives $3(1)+B=4 \implies B=1$.

Substituting $A=1, B=1$ into $A+B+C=3$ gives $1+1+C=3 \implies C=1$.

So, the general term is $a_n = n^2 + n + 1$.

---

Step 2: Find the sum of the first n terms ($S_n$).

$S_n = \sum\limits_{k=1}^{n} a_k = \sum\limits_{k=1}^{n} (k^2 + k + 1)$

$S_n = \sum\limits_{k=1}^{n} k^2 + \sum\limits_{k=1}^{n} k + \sum\limits_{k=1}^{n} 1$

Using the standard summation formulas:

$\sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum\limits_{k=1}^{n} k = \frac{n(n+1)}{2}$

$\sum\limits_{k=1}^{n} 1 = n$

Now, substitute these into the expression for $S_n$:

$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + n$

Take $n$ as a common factor:

$S_n = n \left[ \frac{(n+1)(2n+1)}{6} + \frac{n+1}{2} + 1 \right]$

Find a common denominator inside the bracket (which is 6):

$S_n = n \left[ \frac{(n+1)(2n+1) + 3(n+1) + 6}{6} \right]$

$S_n = \frac{n}{6} [ (2n^2 + 3n + 1) + (3n + 3) + 6 ]

$S_n = \frac{n}{6} [ 2n^2 + 6n + 10 ]$

$S_n = \frac{n \cdot 2(n^2 + 3n + 5)}{6}$

$S_n = \frac{n(n^2 + 3n + 5)}{3}$

---

The sum of the first $n$ terms of the series is $\frac{n(n^2 + 3n + 5)}{3}$.

Given:

$S_1 = \sum\limits_{k=1}^{n} k = \frac{n(n+1)}{2}$ (Sum of first n natural numbers)

$S_2 = \sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ (Sum of squares of first n natural numbers)

$S_3 = \sum\limits_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$ (Sum of cubes of first n natural numbers)

---

To Show:

$9S_2^2 = S_3 (1 + 8S_1)$

---

Proof:

We will evaluate the Left Hand Side (LHS) and the Right Hand Side (RHS) separately and show they are equal.

Left Hand Side (LHS):

LHS = $9S_2^2$

Substitute the formula for $S_2$:

LHS = $9 \left( \frac{n(n+1)(2n+1)}{6} \right)^2$

LHS = $9 \left( \frac{n^2(n+1)^2(2n+1)^2}{36} \right)$

LHS = $\frac{9}{36} \cdot n^2(n+1)^2(2n+1)^2$

---

Right Hand Side (RHS):

RHS = $S_3 (1 + 8S_1)$

Substitute the formulas for $S_3$ and $S_1$:

RHS = $\left( \frac{n(n+1)}{2} \right)^2 \left( 1 + 8 \left( \frac{n(n+1)}{2} \right) \right)$

RHS = $\frac{n^2(n+1)^2}{4} \left( 1 + 4n(n+1) \right)$

RHS = $\frac{n^2(n+1)^2}{4} (1 + 4n^2 + 4n)$

The expression $(4n^2 + 4n + 1)$ is the perfect square of $(2n+1)$.

RHS = $\frac{n^2(n+1)^2}{4} (2n+1)^2$

---

From equations (i) and (ii), we can see that LHS = RHS.

Hence Proved.

Given:

The series $\frac{1^{3}}{1} + \frac{1^{3}\;+\;2^{3}}{1\;+\;3} + \frac{1^{3}\;+\;2^{3}\;+\;3^{3}}{1\;+\;3\;+\;5}+...$

---

To Find:

The sum of the first $n$ terms of the series.

---

Solution:

Step 1: Find the general k-th term ($a_k$) of the series.

The k-th term $a_k$ has a numerator and a denominator.

Numerator of $a_k$ = $1^3 + 2^3 + \dots + k^3 = \sum\limits_{i=1}^{k} i^3 = \left(\frac{k(k+1)}{2}\right)^2 = \frac{k^2(k+1)^2}{4}$.

Denominator of $a_k$ = $1 + 3 + 5 + \dots$ to $k$ terms. This is the sum of the first $k$ odd numbers, which is equal to $k^2$.

Therefore, the k-th term is:

$a_k = \frac{\text{Numerator}}{\text{Denominator}} = \frac{k^2(k+1)^2/4}{k^2} = \frac{(k+1)^2}{4}$

---

Step 2: Find the sum of the first n terms ($S_n$).

$S_n = \sum\limits_{k=1}^{n} a_k = \sum\limits_{k=1}^{n} \frac{(k+1)^2}{4}$

$S_n = \frac{1}{4} \sum\limits_{k=1}^{n} (k^2 + 2k + 1)$

$S_n = \frac{1}{4} \left( \sum\limits_{k=1}^{n} k^2 + 2\sum\limits_{k=1}^{n} k + \sum\limits_{k=1}^{n} 1 \right)$

Using the standard summation formulas:

$S_n = \frac{1}{4} \left[ \frac{n(n+1)(2n+1)}{6} + 2\left(\frac{n(n+1)}{2}\right) + n \right]$

$S_n = \frac{1}{4} \left[ \frac{n(n+1)(2n+1)}{6} + n(n+1) + n \right]$

Take $n$ as a common factor from the terms inside the bracket:

$S_n = \frac{n}{4} \left[ \frac{(n+1)(2n+1)}{6} + (n+1) + 1 \right]$

Find a common denominator of 6 inside the bracket:

$S_n = \frac{n}{4} \left[ \frac{(2n^2+3n+1) + 6(n+1) + 6}{6} \right]$

$S_n = \frac{n}{24} [ 2n^2 + 3n + 1 + 6n + 6 + 6 ]

$S_n = \frac{n(2n^2 + 9n + 13)}{24}$

---

The sum of the series up to $n$ terms is $\frac{n(2n^2 + 9n + 13)}{24}$.

To Show:

$\frac{\sum\limits_{k=1}^{n} k(k+1)^2}{\sum\limits_{k=1}^{n} k^2(k+1)} = \frac{3n + 5}{3n + 1}$

---

Proof:

We will find the sum of the series in the numerator and the denominator separately.

Numerator Sum ($S_N$):

The general term is $T_k = k(k+1)^2 = k(k^2+2k+1) = k^3 + 2k^2 + k$.

$S_N = \sum\limits_{k=1}^{n} (k^3 + 2k^2 + k) = \sum\limits k^3 + 2\sum\limits k^2 + \sum\limits k$

$S_N = \frac{n^2(n+1)^2}{4} + 2\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$

Factor out the common term $\frac{n(n+1)}{12}$:

$S_N = \frac{n(n+1)}{12} [3n(n+1) + 4(2n+1) + 6]$

$S_N = \frac{n(n+1)}{12} [3n^2 + 3n + 8n + 4 + 6]$

$S_N = \frac{n(n+1)}{12} [3n^2 + 11n + 10]$

Factoring the quadratic gives $3n^2 + 11n + 10 = (n+2)(3n+5)$.

$S_N = \frac{n(n+1)(n+2)(3n+5)}{12}$

---

Denominator Sum ($S_D$):

The general term is $T_k = k^2(k+1) = k^3 + k^2$.

$S_D = \sum\limits_{k=1}^{n} (k^3 + k^2) = \sum\limits k^3 + \sum\limits k^2$

$S_D = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{6}$

Factor out the common term $\frac{n(n+1)}{12}$:

$S_D = \frac{n(n+1)}{12} [3n(n+1) + 2(2n+1)]$

$S_D = \frac{n(n+1)}{12} [3n^2 + 3n + 4n + 2]$

$S_D = \frac{n(n+1)}{12} [3n^2 + 7n + 2]$

Factoring the quadratic gives $3n^2 + 7n + 2 = (n+2)(3n+1)$.

$S_D = \frac{n(n+1)(n+2)(3n+1)}{12}$

---

The Ratio:

Now we compute the ratio of the two sums:

$\frac{S_N}{S_D} = \frac{\frac{n(n+1)(n+2)(3n+5)}{12}}{\frac{n(n+1)(n+2)(3n+1)}{12}}$

Cancelling the common factor $\frac{n(n+1)(n+2)}{12}$ from the numerator and denominator, we get:

$\frac{S_N}{S_D} = \frac{3n+5}{3n+1}$

Hence Proved.

Given:

Cost of the tractor = $\textsf{₹ } 12000$

Cash payment = $\textsf{₹ } 6000$

Annual principal instalment = $\textsf{₹ } 500$

Interest rate = 12% per annum on the unpaid amount.

---

To Find:

The total cost of the tractor (including cash payment and total interest).

---

Solution:

The balance amount to be paid in instalments is:

Balance Amount = Cost of Tractor - Cash Payment

Balance Amount = $\textsf{₹ } 12000 - \textsf{₹ } 6000 = \textsf{₹ } 6000$.

The balance amount is paid in annual principal instalments of $\textsf{₹ } 500$.

Number of instalments = $\frac{\text{Balance Amount}}{\text{Annual Principal Instalment}}$

Number of instalments = $\frac{6000}{500} = 12$ instalments.

Interest is paid at 12% per annum on the unpaid amount at the beginning of each year.

The unpaid amounts at the beginning of each year (before the principal payment for that year) are:

Year 1: $\textsf{₹ } 6000$

Year 2: $\textsf{₹ } 6000 - \textsf{₹ } 500 = \textsf{₹ } 5500$

Year 3: $\textsf{₹ } 5500 - \textsf{₹ } 500 = \textsf{₹ } 5000$

$\dots$

Year 12: $\textsf{₹ } 500$ (after 11 payments of $\textsf{₹ } 500$ each, the unpaid amount is $6000 - 11 \times 500 = 6000 - 5500 = 500$).

The unpaid amounts form an Arithmetic Progression:

$6000, 5500, 5000, \dots, 500$.

This A.P. has the first term $a = 6000$, common difference $d = -500$, and the number of terms $n = 12$. The last term is $l = 500$.

The total interest paid is 12% of the sum of these unpaid amounts over the 12 years.

Sum of Unpaid Amounts ($S_{12}$) = $\frac{n}{2}(a+l)$

$S_{12} = \frac{12}{2}(6000 + 500) = 6(6500) = \textsf{₹ } 39000$.

Total Interest Paid = 12% of the Sum of Unpaid Amounts

Total Interest Paid = $\frac{12}{100} \times 39000$

Total Interest Paid = $12 \times 390 = \textsf{₹ } 4680$.

The total cost of the tractor is the sum of the cash payment, the balance principal paid, and the total interest paid.

Total Cost = Cash Payment + Balance Amount + Total Interest Paid

Total Cost = $\textsf{₹ } 6000 + \textsf{₹ } 6000 + \textsf{₹ } 4680$

Total Cost = $\textsf{₹ } 12000 + \textsf{₹ } 4680 = \textsf{₹ } 16680$.

The tractor will cost him $\textsf{₹ } 16680$.

Given:

Cost of the scooter = $\textsf{₹ } 22000$

Cash payment = $\textsf{₹ } 4000$

Annual principal instalment = $\textsf{₹ } 1000$

Interest rate = 10% per annum on the unpaid amount.

---

To Find:

The total cost of the scooter (including cash payment and total interest).

---

Solution:

The balance amount to be paid in instalments is:

Balance Amount = Cost of Scooter - Cash Payment

Balance Amount = $\textsf{₹ } 22000 - \textsf{₹ } 4000 = \textsf{₹ } 18000$.

The balance amount is paid in annual principal instalments of $\textsf{₹ } 1000$.

Number of instalments = $\frac{\text{Balance Amount}}{\text{Annual Principal Instalment}}$

Number of instalments = $\frac{18000}{1000} = 18$ instalments.

Interest is paid at 10% per annum on the unpaid amount at the beginning of each year.

The unpaid amounts at the beginning of each year (before the principal payment for that year) are:

Year 1: $\textsf{₹ } 18000$

Year 2: $\textsf{₹ } 18000 - \textsf{₹ } 1000 = \textsf{₹ } 17000$

Year 3: $\textsf{₹ } 17000 - \textsf{₹ } 1000 = \textsf{₹ } 16000$

$\dots$

Year 18: $\textsf{₹ } 18000 - (17 \times \textsf{₹ } 1000) = \textsf{₹ } 18000 - \textsf{₹ } 17000 = \textsf{₹ } 1000$.

The unpaid amounts form an Arithmetic Progression:

$18000, 17000, 16000, \dots, 1000$.

This A.P. has the first term $a = 18000$, common difference $d = -1000$, and the number of terms $n = 18$. The last term is $l = 1000$.

The total interest paid is 10% of the sum of these unpaid amounts over the 18 years.

Sum of Unpaid Amounts ($S_{18}$) = $\frac{n}{2}(a+l)$

$S_{18} = \frac{18}{2}(18000 + 1000) = 9(19000) = \textsf{₹ } 171000$.

Total Interest Paid = 10% of the Sum of Unpaid Amounts

Total Interest Paid = $\frac{10}{100} \times 171000$

Total Interest Paid = $\frac{1}{10} \times 171000 = \textsf{₹ } 17100$.

The total cost of the scooter is the sum of the cash payment, the balance principal paid, and the total interest paid.

Total Cost = Cash Payment + Balance Amount + Total Interest Paid

Total Cost = $\textsf{₹ } 4000 + \textsf{₹ } 18000 + \textsf{₹ } 17100$

Total Cost = $\textsf{₹ } 22000 + \textsf{₹ } 17100 = \textsf{₹ } 39100$.

The scooter will cost him $\textsf{₹ } 39100$.

Given:

A person starts a letter chain by sending letters to 4 friends.

Each recipient sends letters to 4 new people.

The cost to mail one letter is 50 paise = $\textsf{₹ } 0.50$.

---

To Find:

The total amount spent on postage up to and including the mailing of the 8th set of letters.

---

Solution:

The number of letters mailed in each set forms a Geometric Progression (G.P.).

Number of letters in the 1st set = 4 = $4^1$

Number of letters in the 2nd set = $4 \times 4 = 16 = 4^2$

Number of letters in the 3rd set = $16 \times 4 = 64 = 4^3$

And so on.

To find the total amount spent when the 8th set is mailed, we need to find the total number of letters sent in all 8 sets.

Total number of letters = Sum of letters in Set 1 to Set 8

Total Letters = $4 + 4^2 + 4^3 + \dots + 4^8$

This is a G.P. with:

First term, $a = 4$

Common ratio, $r = 4$

Number of terms, $n = 8$

The sum of the first $n$ terms of a G.P. is given by the formula $S_n = \frac{a(r^n - 1)}{r-1}$.

Total Letters, $S_8 = \frac{4(4^8 - 1)}{4 - 1}$

First, we calculate $4^8$: $4^8 = (2^2)^8 = 2^{16} = 65536$.

$S_8 = \frac{4(65536 - 1)}{3}$

$S_8 = \frac{4 \times 65535}{3}$

$S_8 = 4 \times 21845$

$S_8 = 87380$

So, a total of 87,380 letters were mailed.

The cost of mailing one letter is 50 paise, which is $\textsf{₹ } 0.50$.

Total amount spent = (Total number of letters) $\times$ (Cost per letter)

Total amount spent = $87380 \times 0.50$

Total amount spent = $\textsf{₹ } 43690$

---

The total amount spent on postage is $\textsf{₹ } 43,690$.

Given:

Principal amount (P) = $\textsf{₹ } 10,000$

Rate of simple interest (R) = 5% per annum.

---

To Find:

1. The amount in the 15th year.

2. The total amount after 20 years.

---

Solution:

First, we calculate the simple interest earned each year.

Annual Simple Interest = $\frac{P \times R \times T}{100}$, where T = 1 year.

Annual Interest = $\frac{10000 \times 5 \times 1}{100} = \textsf{₹ } 500$

The amount in the bank increases by a constant $\textsf{₹ } 500$ each year, forming an Arithmetic Progression.

---

1. Amount in the 15th year

The phrase "amount in the 15th year" refers to the amount at the beginning of the 15th year. This is equivalent to the amount at the end of 14 years.

Total interest earned in 14 years = (Annual Interest) $\times$ 14

Total interest after 14 years = $\textsf{₹ } 500 \times 14 = \textsf{₹ } 7,000$

Amount at the beginning of 15th year = Principal + Total Interest after 14 years

Amount = $\textsf{₹ } 10,000 + \textsf{₹ } 7,000 = \textsf{₹ } 17,000$

---

2. Total amount after 20 years

"After 20 years" means at the end of the 20-year period.

Total interest earned in 20 years = (Annual Interest) $\times$ 20

Total interest after 20 years = $\textsf{₹ } 500 \times 20 = \textsf{₹ } 10,000$

Total amount after 20 years = Principal + Total Interest after 20 years

Total amount = $\textsf{₹ } 10,000 + \textsf{₹ } 10,000 = \textsf{₹ } 20,000$

---

Thus, the amount in the 15th year is $\textsf{₹ } 17,000$ and the total amount after 20 years is $\textsf{₹ } 20,000$.

Given:

Initial cost of the machine, $P = \textsf{₹ } 15,625$

Annual depreciation rate, $r = 20\% = 0.20$

Time period, $n = 5$ years.

---

To Find:

The estimated value of the machine after 5 years.

---

Solution:

The value of the machine decreases each year by a fixed percentage. This forms a Geometric Progression.

The formula for the depreciated value ($V$) after $n$ years is:

$V = P(1 - r)^n$

Substitute the given values into the formula:

$V = 15625(1 - 0.20)^5$

$V = 15625(0.8)^5$

We can write $0.8$ as the fraction $\frac{4}{5}$:

$V = 15625 \left(\frac{4}{5}\right)^5$

$V = 15625 \times \frac{4^5}{5^5}$

$V = 15625 \times \frac{1024}{3125}$

Now, we can simplify the expression. We notice that $15625 = 125^2$ and $3125 = 5^5$. Alternatively, divide 15625 by 3125:

$\frac{15625}{3125} = 5$

So, the expression becomes:

$V = 5 \times 1024$

$V = 5120$

---

The estimated value of the machine at the end of 5 years is $\textsf{₹ } 5,120$.

Given:

Initial number of workers = 150.

From the second day onwards, 4 workers dropped out each day.

The work was finished 8 days later than planned.

---

To Find:

The actual number of days in which the work was completed.

---

Solution:

Let the originally planned number of days to finish the work be $d$.

The total amount of work is measured in "man-days".

Planned Scenario:

Total Work = (Number of workers) $\times$ (Number of days)

Actual Scenario:

The work took 8 more days than planned, so the actual number of days is $d+8$.

The number of workers each day is as follows:

Day 1: 150 workers

Day 2: 150 - 4 = 146 workers

Day 3: 146 - 4 = 142 workers

This forms an Arithmetic Progression (A.P.) where the first term is $a=150$, the common difference is $d'=-4$, and the number of terms is $n = d+8$.

The total work done is the sum of this A.P. The formula for the sum of an A.P. is $S_n = \frac{n}{2}[2a + (n-1)d']$.

Total Work = $S_{d+8} = \frac{d+8}{2}[2(150) + ((d+8)-1)(-4)]$

Total Work = $\frac{d+8}{2}[300 + (d+7)(-4)]$

Total Work = $\frac{d+8}{2}[300 - 4d - 28]$

Total Work = $\frac{d+8}{2}[272 - 4d]$

Since the work done is the same in both scenarios, we equate (i) and (ii):

$150d = (d+8)(136 - 2d)$

$150d = 136d - 2d^2 + 1088 - 16d$

$150d = 120d - 2d^2 + 1088$

Rearranging into a standard quadratic equation form ($ax^2+bx+c=0$):

$2d^2 + 150d - 120d - 1088 = 0$

$2d^2 + 30d - 1088 = 0$

Dividing by 2:

$d^2 + 15d - 544 = 0$

Solving the quadratic equation by factorization:

$d^2 + 32d - 17d - 544 = 0$

$d(d+32) - 17(d+32) = 0$

$(d-17)(d+32) = 0$

The possible values for $d$ are $17$ and $-32$. Since the number of days cannot be negative, the planned number of days was $d = 17$.

The question asks for the actual number of days in which the work was completed.

Actual number of days = $d + 8 = 17 + 8 = 25$

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The number of days in which the work was completed is 25.