Chapter 8 Sequences And Series

Given:

Two sequences defined by the explicit formulas:

(i) $a_n = 2n + 5$

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

To Find:

The first three terms ($a_1, a_2, a_3$) for each sequence.

Solution:

For part (i), the $n$-th term is given by $a_n = 2n + 5$.

To find the first term, we substitute $n=1$:

$a_1 = 2(1) + 5 = 2 + 5 = 7$

To find the second term, we substitute $n=2$:

$a_2 = 2(2) + 5 = 4 + 5 = 9$

To find the third term, we substitute $n=3$:

$a_3 = 2(3) + 5 = 6 + 5 = 11$

The first three terms of the sequence defined by $a_n = 2n + 5$ are 7, 9, and 11.

For part (ii), the $n$-th term is given by $a_n = \frac{n - 3}{4}$.

To find the first term, we substitute $n=1$:

$a_1 = \frac{1 - 3}{4} = \frac{-2}{4} = -\frac{1}{2}$

To find the second term, we substitute $n=2$:

$a_2 = \frac{2 - 3}{4} = \frac{-1}{4}$

To find the third term, we substitute $n=3$:

$a_3 = \frac{3 - 3}{4} = \frac{0}{4} = 0$

The first three terms of the sequence defined by $a_n = \frac{n - 3}{4}$ are $-\frac{1}{2}$, $-\frac{1}{4}$, and 0.

Given:

The sequence is defined by the formula for the $n$-th term: $a_n = (n – 1) (2 – n) (3 + n)$.

To Find:

The 20^th term of the sequence, which is $a_{20}$.

Solution:

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

$a_{20} = (20 – 1) (2 – 20) (3 + 20)$

Calculate the value of each factor in the expression:

$20 - 1 = 19$

$2 - 20 = -18$

$3 + 20 = 23$

Now, substitute these values back into the expression for $a_{20}$:

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

Multiply the terms:

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

First, multiply 19 by -18:

$19 \times (-18) = -(19 \times 18)$

$19 \times 18 = 342$

So, $19 \times (-18) = -342$.

Now, multiply -342 by 23:

$a_{20} = -342 \times 23 = -7866$.

The 20^th term of the sequence is -7866.

Given:

A sequence defined by a recursive relation:

$a_1 = 1$

$a_n = a_{n – 1} + 2$ for $n \geq 2$

To Find:

The terms of the sequence defined by the given recurrence relation. We will find the first few terms to understand the sequence.

Solution:

The first term is given directly:

$a_1 = 1$

To find the second term, we use the recurrence relation with $n=2$:

$a_2 = a_{2-1} + 2 = a_1 + 2$

Substitute the value of $a_1$:

To find the third term, we use the recurrence relation with $n=3$:

$a_3 = a_{3-1} + 2 = a_2 + 2$

Substitute the value of $a_2$:

To find the fourth term, we use the recurrence relation with $n=4$:

$a_4 = a_{4-1} + 2 = a_3 + 2$

Substitute the value of $a_3$:

The sequence starts with 1, 3, 5, 7, ...

This sequence is an arithmetic progression with the first term $a_1 = 1$ and a common difference of $d = 2$.

To find the first five terms of the sequence, we need to substitute $n=1, 2, 3, 4, 5$ in the given formula for the $n^{th}$ term.

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

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

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

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

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

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

To find the first five terms of the sequence, we need to substitute $n=1, 2, 3, 4, 5$ in the given formula for the $n^{th}$ term.

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

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

To find the first five terms of the sequence, we need to substitute $n=1, 2, 3, 4, 5$ in the given formula for the $n^{th}$ term.

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

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

To find the first five terms of the sequence, we need to substitute $n=1, 2, 3, 4, 5$ in the given formula for the $n^{th}$ term.

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 $n=3$: $a_3 = \frac{2(3) - 3}{6} = \frac{6 - 3}{6} = \frac{3}{6} = \frac{\cancel{3}^{1}}{\cancel{6}_{2}} = \frac{1}{2}$

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

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

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

To find the first five terms of the sequence, we need to substitute $n=1, 2, 3, 4, 5$ in the given formula for the $n^{th}$ term.

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

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

To find the first five terms of the sequence, we need to substitute $n=1, 2, 3, 4, 5$ in the given formula for the $n^{th}$ term.

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

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

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

For $n=4$: $a_4 = 4 \times \frac{4^2+5}{4} = 4 \times \frac{16+5}{4} = 4 \times \frac{21}{4} = \frac{84}{4} = 21$

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

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

Given:

To Find:

The $17^{th}$ term ($a_{17}$) and the $24^{th}$ term ($a_{24}$).

Solution:

To find the $17^{th}$ term, substitute $n=17$ in the formula for $a_n$:

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

$a_{17} = 68 - 3$

$a_{17} = 65$

To find the $24^{th}$ term, substitute $n=24$ in the formula for $a_n$:

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

$a_{24} = 96 - 3$

$a_{24} = 93$

Thus, the $17^{th}$ term is 65 and the $24^{th}$ term is 93.

Given:

To Find:

The $7^{th}$ term ($a_7$).

Solution:

To find the $7^{th}$ term, substitute $n=7$ in the formula for $a_n$:

$a_7 = \frac{7^2}{2^7}$

$a_7 = \frac{49}{128}$

Thus, the $7^{th}$ term of the sequence is $\frac{49}{128}$.

Given:

To Find:

The $9^{th}$ term ($a_9$).

Solution:

To find the $9^{th}$ term, substitute $n=9$ in the formula for $a_n$:

$a_9 = (-1)^{9-1} 9^3$

$a_9 = (-1)^8 9^3$

Since $(-1)^8 = 1$: $a_9 = 1 \times 9^3$

Calculate $9^3$: $9^3 = 9 \times 9 \times 9 = 81 \times 9 = 729$

Therefore: $a_9 = 1 \times 729 = 729$

Thus, the $9^{th}$ term of the sequence is 729.

Given:

To Find:

The $20^{th}$ term ($a_{20}$).

Solution:

To find the $20^{th}$ term, substitute $n=20$ in the formula for $a_n$:

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

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

Calculate the numerator: $20 \times 18 = 360$

Calculate the denominator: $20 + 3 = 23$

Substitute these values back into the expression for $a_{20}$: $a_{20} = \frac{360}{23}$

Thus, the $20^{th}$ term of the sequence is $\frac{360}{23}$.

We are given the first term of the sequence and a recurrence relation to find the subsequent terms. We need to find the first five terms and the corresponding series.

Given:

To Find:

The first five terms of the sequence and the corresponding series.

Solution:

The first term is already given: $a_1 = 3$

Using the recurrence relation $a_n = 3a_{n-1} + 2$, we find the terms for $n=2, 3, 4, 5$.

For $n=2$: $a_2 = 3a_{2-1} + 2 = 3a_1 + 2 = 3(3) + 2 = 9 + 2 = 11$

For $n=3$: $a_3 = 3a_{3-1} + 2 = 3a_2 + 2 = 3(11) + 2 = 33 + 2 = 35$

For $n=4$: $a_4 = 3a_{4-1} + 2 = 3a_3 + 2 = 3(35) + 2 = 105 + 2 = 107$

For $n=5$: $a_5 = 3a_{5-1} + 2 = 3a_4 + 2 = 3(107) + 2 = 321 + 2 = 323$

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

Corresponding Series:

The corresponding series is the sum of the first five terms:

Series = $a_1 + a_2 + a_3 + a_4 + a_5$

Series = $3 + 11 + 35 + 107 + 323$

Calculating the sum:

$3 + 11 = 14$

$14 + 35 = 49$

$49 + 107 = 156$

$156 + 323 = 479$

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

The corresponding series is $3 + 11 + 35 + 107 + 323 = 479$.

We are given the first term of the sequence and a recurrence relation to find the subsequent terms. We need to find the first five terms and the corresponding series.

Given:

To Find:

The first five terms of the sequence and the corresponding series.

Solution:

The first term is already given: $a_1 = -1$

Using the recurrence relation $a_n = \frac{a_{n-1}}{n}$, we find the terms for $n=2, 3, 4, 5$.

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

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

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

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

The first five terms of the sequence are $-1, \frac{-1}{2}, \frac{-1}{6}, \frac{-1}{24},$ and $\frac{-1}{120}$.

Corresponding Series:

The corresponding series is the sum of the first five terms:

Series = $a_1 + a_2 + a_3 + a_4 + a_5$

Series = $-1 + \left(\frac{-1}{2}\right) + \left(\frac{-1}{6}\right) + \left(\frac{-1}{24}\right) + \left(\frac{-1}{120}\right)$

To add these fractions, we find a common denominator, which is the LCM of 1, 2, 6, 24, and 120.

LCM(1, 2, 6, 24, 120) = 120.

Series = $\frac{-1 \times 120}{1 \times 120} + \frac{-1 \times 60}{2 \times 60} + \frac{-1 \times 20}{6 \times 20} + \frac{-1 \times 5}{24 \times 5} + \frac{-1}{120}$

Series = $\frac{-120}{120} + \frac{-60}{120} + \frac{-20}{120} + \frac{-5}{120} + \frac{-1}{120}$

Series = $\frac{-120 - 60 - 20 - 5 - 1}{120}$

Series = $\frac{-206}{120}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2.

$\frac{\cancel{-206}^{-103}}{\cancel{120}_{60}}$

Series = $\frac{-103}{60}$

The first five terms of the sequence are $-1, \frac{-1}{2}, \frac{-1}{6}, \frac{-1}{24}, \frac{-1}{120}$.

The corresponding series is $-1 + \left(\frac{-1}{2}\right) + \left(\frac{-1}{6}\right) + \left(\frac{-1}{24}\right) + \left(\frac{-1}{120}\right) = \frac{-103}{60}$.

We are given the first two terms of the sequence and a recurrence relation to find the subsequent terms. We need to find the first five terms and the corresponding series.

Given:

To Find:

The first five terms of the sequence and the corresponding series.

Solution:

The first two terms are already given:

$a_1 = 2$

$a_2 = 2$

Using the recurrence relation $a_n = a_{n-1} - 1$, we find the terms for $n=3, 4, 5$.

For $n=3$: $a_3 = a_{3-1} - 1 = a_2 - 1 = 2 - 1 = 1$

For $n=4$: $a_4 = a_{4-1} - 1 = a_3 - 1 = 1 - 1 = 0$

For $n=5$: $a_5 = a_{5-1} - 1 = a_4 - 1 = 0 - 1 = -1$

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

Corresponding Series:

The corresponding series is the sum of the first five terms:

Series = $a_1 + a_2 + a_3 + a_4 + a_5$

Series = $2 + 2 + 1 + 0 + (-1)$

Calculating the sum:

$2 + 2 = 4$

$4 + 1 = 5$

$5 + 0 = 5$

$5 + (-1) = 4$

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

The corresponding series is $2 + 2 + 1 + 0 + (-1) = 4$.

We are given the definition of the Fibonacci sequence. We need to find the first few terms and then calculate the specified ratios.

Given:

To Find:

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

Solution:

First, let's find the first six terms of the Fibonacci sequence.

$a_1 = 1$

$a_2 = 1$

For $n=3$: $a_3 = a_{3-1} + a_{3-2} = a_2 + a_1 = 1 + 1 = 2$

For $n=4$: $a_4 = a_{4-1} + a_{4-2} = a_3 + a_2 = 2 + 1 = 3$

For $n=5$: $a_5 = a_{5-1} + a_{5-2} = a_4 + a_3 = 3 + 2 = 5$

For $n=6$: $a_6 = a_{6-1} + a_{6-2} = a_5 + a_4 = 5 + 3 = 8$

So the first six terms are 1, 1, 2, 3, 5, 8.

Now, we calculate the ratios $\frac{a_{n+1}}{a_n}$ for $n=1, 2, 3, 4, 5$.

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

Thus, the values of $\frac{a_{n+1}}{a_n}$ for $n=1, 2, 3, 4, 5$ are $1, 2, \frac{3}{2}, \frac{5}{3},$ and $\frac{8}{5}$, respectively.

Given:

To Find:

The $10^{th}$ term ($a_{10}$) and the $n^{th}$ term ($a_n$) of the G.P.

Solution:

In the given G.P., the first term is $a_1 = 5$.

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

$r = \frac{a_2}{a_1} = \frac{25}{5} = 5$

Also, $r = \frac{a_3}{a_2} = \frac{125}{25} = 5$.

So, the first term is $a = 5$ and the common ratio is $r = 5$.

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

To find the $n^{th}$ term, substitute the values of $a$ and $r$ into the formula:

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

Using the property of exponents $x^m \times x^n = x^{m+n}$: $a_n = 5^{1} \times 5^{n-1} = 5^{1 + (n-1)} = 5^n$

To find the $10^{th}$ term, substitute $n=10$ into the formula for the $n^{th}$ term (i):

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

Calculating $5^{10}$:

$5^{10} = 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5$

$5^{10} = 9765625$

Thus, the $10^{th}$ term of the G.P. is $5^{10}$ or 9765625 and the $n^{th}$ term is $5^n$.

Given:

To Find:

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

Solution:

In the given G.P., the first term is $a = 2$.

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

$r = \frac{8}{2} = 4$

Also, $r = \frac{32}{8} = 4$.

So, the first term is $a = 2$ and the common ratio is $r = 4$.

Let the $n^{th}$ term of the G.P. be $a_n$. The formula for the $n^{th}$ term of a G.P. is:

We are given that the $n^{th}$ term is $131072$. Substitute the values of $a_n$, $a$, and $r$ into equation (i):

$131072 = 2 \times 4^{n-1}$

Divide both sides by 2: $\frac{131072}{2} = 4^{n-1}$

$65536 = 4^{n-1}$

We need to express $65536$ as a power of $4$.

We can calculate powers of 4: $4^1 = 4$ $4^2 = 16$ $4^3 = 64$ $4^4 = 256$ $4^5 = 1024$ $4^6 = 4096$ $4^7 = 16384$ $4^8 = 65536$

So, we have: $65536 = 4^8$

Substitute this back into the equation: $4^8 = 4^{n-1}$

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

$8 = n - 1$

Solve for $n$: $n = 8 + 1$

$n = 9$

Thus, the term $131072$ is the $9^{th}$ term of the given G.P.

Given:

To Find:

The 10$^{th}$ term of the G.P., $a_{10}$.

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 $a_n = ar^{n-1}$.

Using the given information, we can write:

So, $ar^2 = 24$.

So, $ar^5 = 192$.

Divide equation (ii) by equation (i):

$\frac{ar^5}{ar^2} = \frac{192}{24}$

$r^{5-2} = 8$

$r^3 = 8$

To find $r$, take the cube root of both sides: $r = \sqrt[3]{8}$

$r = 2$

Substitute the value of $r=2$ into equation (i):

$a(2)^2 = 24$

$4a = 24$

Divide both sides by 4: $a = \frac{24}{4}$

$a = 6$

So, the first term is 6 and the common ratio is 2.

Now, we need to find the 10$^{th}$ term, $a_{10}$, using the formula $a_n = ar^{n-1}$ with $n=10$.

$a_{10} = ar^{10-1} = ar^9$

Substitute the values $a=6$ and $r=2$: $a_{10} = 6 \times 2^9$

Calculate $2^9$: $2^9 = 512$

Substitute this value back into the expression for $a_{10}$: $a_{10} = 6 \times 512$

Calculate the product: $6 \times 512 = 3072$

So, $a_{10} = 3072$.

Thus, the 10$^{th}$ term of the G.P. is 3072.

Given:

To Find:

The sum of the first $n$ terms ($S_n$) and the sum of the first 5 terms ($S_5$).

Solution:

The given series is a Geometric Progression (G.P.).

The first term is $a = 1$.

The common ratio, $r$, is the ratio of any term to its preceding term:

$r = \frac{\text{2nd term}}{\text{1st term}} = \frac{2/3}{1} = \frac{2}{3}$

Alternatively, $r = \frac{\text{3rd term}}{\text{2nd term}} = \frac{4/9}{2/3} = \frac{4}{9} \times \frac{3}{2} = \frac{12}{18} = \frac{2}{3}$.

Since $|r| = |\frac{2}{3}| < 1$, the formula for the sum of the first $n$ terms of a G.P. is given by $S_n = a \frac{1 - r^n}{1 - r}$.

Substitute the values of $a=1$ and $r=\frac{2}{3}$ into the formula for $S_n$:

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

Simplify the denominator: $1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

Substitute this back into the expression for $S_n$: $S_n = \frac{1 - (\frac{2}{3})^n}{\frac{1}{3}}$

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

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

$S_n = 3 - 3 \times \frac{2^n}{3^n}$

$S_n = 3 - \frac{3^1 \times 2^n}{3^n}$

Using the property of exponents $\frac{x^m}{x^n} = x^{m-n}$: $S_n = 3 - 2^n \times 3^{1-n}$

Or, $S_n = 3 - \frac{2^n}{3^{n-1}}$.

Now, we need to find the sum of the first 5 terms ($S_5$). Substitute $n=5$ into equation (i):

$S_5 = 3 - \frac{2^5}{3^{5-1}}$

$S_5 = 3 - \frac{2^5}{3^4}$

Calculate $2^5$ and $3^4$: $2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$

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

Substitute these values back into the expression for $S_5$: $S_5 = 3 - \frac{32}{81}$

To subtract the fraction, find a common denominator (81): $S_5 = \frac{3 \times 81}{81} - \frac{32}{81}$

$S_5 = \frac{243}{81} - \frac{32}{81}$

$S_5 = \frac{243 - 32}{81}$

$S_5 = \frac{211}{81}$

Thus, the sum of the first $n$ terms is $3 - \frac{2^n}{3^{n-1}}$ and the sum of the first 5 terms is $\frac{211}{81}$.

Given:

To Find:

The number of terms ($n$) required to give the sum $\frac{3069}{512}$.

Solution:

In the given G.P., the first term is $a = 3$.

The common ratio, $r$, is the ratio of any term to its preceding term:

$r = \frac{\text{2nd term}}{\text{1st term}} = \frac{3/2}{3} = \frac{3}{2} \times \frac{1}{3} = \frac{\cancel{3}}{\cancel{3} \times 2} = \frac{1}{2}$

Since $|r| = |\frac{1}{2}| < 1$, the formula for the sum of the first $n$ terms of a G.P. is given by:

We are given $S_n = \frac{3069}{512}$, $a=3$, and $r=\frac{1}{2}$. Substitute these values into equation (i):

$\frac{3069}{512} = 3 \frac{1 - (\frac{1}{2})^n}{1 - \frac{1}{2}}$

Simplify the denominator $1 - \frac{1}{2}$: $1 - \frac{1}{2} = \frac{2}{2} - \frac{1}{2} = \frac{1}{2}$

Substitute the denominator value back into the equation: $\frac{3069}{512} = 3 \frac{1 - (\frac{1}{2})^n}{\frac{1}{2}}$

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

Divide both sides by 6: $\frac{3069}{512 \times 6} = 1 - \left(\frac{1}{2}\right)^n$

$\frac{3069}{3072} = 1 - \left(\frac{1}{2}\right)^n$

Simplify the fraction $\frac{3069}{3072}$ by dividing the numerator and denominator by 3: $\frac{3069 \div 3}{3072 \div 3} = \frac{1023}{1024}$

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

Rearrange the equation to solve for $(\frac{1}{2})^n$: $\left(\frac{1}{2}\right)^n = 1 - \frac{1023}{1024}$

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

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

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

Express the right side as a power of $\frac{1}{2}$. We know that $1024 = 2^{10}$. $\frac{1}{1024} = \frac{1}{2^{10}} = \left(\frac{1}{2}\right)^{10}$

So the equation becomes: $\left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^{10}$

Since the bases are equal, the exponents must be equal: $n = 10$

Thus, 10 terms of the G.P. are needed to give the sum $\frac{3069}{512}$.

Given:

Sum of the first three terms of a G.P. $= \frac{13}{12}$

Product of the first three terms of a G.P. $= -1$

To Find:

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

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.

According to the problem, the product of the first three terms is $-1$.

$(\frac{a}{r})(a)(ar) = -1$

$a^3 = -1$

Taking the cube root of both sides, we get:

$a = -1$

According to the problem, the sum of the first three terms is $\frac{13}{12}$.

$\frac{a}{r} + a + ar = \frac{13}{12}$

Substitute the value of $a = -1$ into this equation:

$\frac{-1}{r} + (-1) + (-1)r = \frac{13}{12}$

$-\frac{1}{r} - 1 - r = \frac{13}{12}$

Multiply the entire equation by $-1$ to make the terms positive:

$\frac{1}{r} + 1 + r = -\frac{13}{12}$

Rearrange the terms to form a quadratic equation in $r$:

$r + \frac{1}{r} + 1 = -\frac{13}{12}$

Multiply the equation by $12r$ to eliminate the fractions:

$12r(r) + 12r(\frac{1}{r}) + 12r(1) = 12r(-\frac{13}{12})$

$12r^2 + 12 + 12r = -13r$

Move all terms to one side to get a standard quadratic equation:

$12r^2 + 12r + 13r + 12 = 0$

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

This is a quadratic equation of the form $Ax^2 + Bx + C = 0$, where $A=12$, $B=25$, and $C=12$. We can solve for $r$ using the quadratic formula $r = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$.

$r = \frac{-(25) \pm \sqrt{(25)^2 - 4(12)(12)}}{2(12)}$

$r = \frac{-25 \pm \sqrt{625 - 576}}{24}$

$r = \frac{-25 \pm \sqrt{49}}{24}$

$r = \frac{-25 \pm 7}{24}$

We have two possible values for the common ratio $r$:

Case 1: $r = \frac{-25 + 7}{24} = \frac{-18}{24} = -\frac{3}{4}$

Case 2: $r = \frac{-25 - 7}{24} = \frac{-32}{24} = -\frac{4}{3}$

Now we find the terms of the G.P. for each case, using $a=-1$.

Case 1: $a = -1$ and $r = -\frac{3}{4}$

First term = $\frac{a}{r} = \frac{-1}{-3/4} = -1 \times (-\frac{4}{3}) = \frac{4}{3}$

Second term = $a = -1$

Third term = $ar = (-1)(-\frac{3}{4}) = \frac{3}{4}$

The terms are $\frac{4}{3}, -1, \frac{3}{4}$.

Case 2: $a = -1$ and $r = -\frac{4}{3}$

First term = $\frac{a}{r} = \frac{-1}{-4/3} = -1 \times (-\frac{3}{4}) = \frac{3}{4}$

Second term = $a = -1$

Third term = $ar = (-1)(-\frac{4}{3}) = \frac{4}{3}$

The terms are $\frac{3}{4}, -1, \frac{4}{3}$.

The terms obtained in both cases are the same, just in reverse order.

Therefore, the common ratio is either $-\frac{3}{4}$ or $-\frac{4}{3}$, and the terms of the G.P. are $\frac{4}{3}, -1,$ and $\frac{3}{4}$.

Final Answer:

The common ratio $r$ is either $-\frac{3}{4}$ or $-\frac{4}{3}$.

The terms of the G.P. are $\frac{4}{3}, -1, \frac{3}{4}$.

Given:

The sequence is 7, 77, 777, 7777, ...

To Find:

The sum of the sequence to $n$ terms.

Solution:

Let the sum of the sequence to $n$ terms be $S_n$.

$S_n = 7 + 77 + 777 + \dots + \underbrace{77\dots7}_{n \text{ terms}}$

We can write each term as a product of 7 and a sequence of 1s.

$S_n = 7(1) + 7(11) + 7(111) + \dots + 7(\underbrace{11\dots1}_{n \text{ terms}})$

Factor out 7:

$S_n = 7 (1 + 11 + 111 + \dots + \underbrace{11\dots1}_{n \text{ terms}})$

Now, we can express the terms within the parenthesis using powers of 10. Note that $\underbrace{11\dots1}_{k \text{ terms}} = \frac{10^k - 1}{9}$.

So, $1 = \frac{10^1 - 1}{9}$, $11 = \frac{10^2 - 1}{9}$, $111 = \frac{10^3 - 1}{9}$, and so on.

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

Factor out $\frac{1}{9}$:

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

Group the terms involving powers of 10 and the $-1$ terms separately:

$S_n = \frac{7}{9} \left[ (10^1 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + 1 \text{ ($n$ times)}) \right]$

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

The expression $10^1 + 10^2 + 10^3 + \dots + 10^n$ is a geometric series with first term $a = 10$, common ratio $r = 10$, and $n$ terms. The sum of this G.P. is given by the formula $S_{\text{G.P.}} = \frac{a(r^n - 1)}{r - 1}$.

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

Substitute this back into the expression for $S_n$:

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

To simplify further, we can find a common denominator inside the bracket:

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

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

$S_n = \frac{7}{81} [10(10^n - 1) - 9n]$

$S_n = \frac{7}{81} [10^{n+1} - 10 - 9n]$

Final Answer:

The sum of the sequence 7, 77, 777, ... to $n$ terms is $S_n = \frac{7}{81} (10^{n+1} - 9n - 10)$.

Given:

Number of ancestors in the 1st preceding generation (parents) = 2

Number of ancestors in the 2nd preceding generation (grandparents) = 4

Number of ancestors in the 3rd preceding generation (great grandparents) = 8

Number of generations to consider = 10

To Find:

The total number of ancestors during the ten generations preceding his own.

Solution:

The number of ancestors in each preceding generation forms the sequence: 2, 4, 8, ...

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

The first term, which represents the number of ancestors in the first preceding generation (parents), is $a = 2$.

The common ratio is the factor by which the number of ancestors increases from one generation to the next. $r = \frac{4}{2} = 2$ or $r = \frac{8}{4} = 2$. So, $r=2$.

We need to find the total number of ancestors over the first 10 such generations. Thus, the number of terms is $n = 10$.

The sum of the first $n$ terms of a G.P. is given by the formula:

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

Substitute the values $a=2$, $r=2$, and $n=10$ into the formula:

$S_{10} = \frac{2(2^{10} - 1)}{2 - 1}$

$S_{10} = \frac{2(2^{10} - 1)}{1}$

$S_{10} = 2(2^{10} - 1)$

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

$2^{10} = 1024$

Now, substitute this value back into the expression for $S_{10}$:

$S_{10} = 2(1024 - 1)$

$S_{10} = 2(1023)$

Perform the multiplication:

$S_{10} = 2046$

Final Answer:

The number of ancestors during the ten generations preceding the person is 2046.

Given:

The first term of the sequence is 1.

The last term of the sequence is 256.

Three numbers are to be inserted between 1 and 256 to form a G.P.

To Find:

The three numbers that, when inserted between 1 and 256, form a G.P.

Solution:

Let the three numbers to be inserted between 1 and 256 be $G_1, G_2, G_3$.

The resulting sequence in G.P. will be $1, G_1, G_2, G_3, 256$.

This sequence has 5 terms.

The first term is $a_1 = 1$.

The fifth term is $a_5 = 256$.

The formula for the $n$-th term of a G.P. is $a_n = a_1 r^{n-1}$, where $r$ is the common ratio.

For the 5th term ($n=5$):

$a_5 = a_1 r^{5-1}$

$256 = 1 \cdot r^4$

$r^4 = 256$

To find $r$, we take the fourth root of 256. Note that since the power is even, there will be both positive and negative real roots.

$r = \pm \sqrt[4]{256}$

We know that $4^4 = 4 \times 4 \times 4 \times 4 = 16 \times 16 = 256$.

So, $\sqrt[4]{256} = 4$.

Thus, the possible values for the common ratio $r$ are $4$ and $-4$.

Now we find the three inserted numbers for each value of $r$. The inserted terms are $G_1 = ar$, $G_2 = ar^2$, and $G_3 = ar^3$, where $a=1$ is the first term of the G.P.

Case 1: Common ratio $r = 4$

$G_1 = 1 \cdot 4 = 4$

$G_2 = 1 \cdot 4^2 = 1 \cdot 16 = 16$

$G_3 = 1 \cdot 4^3 = 1 \cdot 64 = 64$

The resulting sequence is 1, 4, 16, 64, 256, which is a G.P. with common ratio 4.

Case 2: Common ratio $r = -4$

$G_1 = 1 \cdot (-4) = -4$

$G_2 = 1 \cdot (-4)^2 = 1 \cdot 16 = 16$

$G_3 = 1 \cdot (-4)^3 = 1 \cdot (-64) = -64$

The resulting sequence is 1, -4, 16, -64, 256, which is a G.P. with common ratio -4.

Final Answer:

The three numbers to be inserted between 1 and 256 to form a G.P. are either 4, 16, 64 or -4, 16, -64.

Given:

Arithmetic Mean (A.M.) of two positive numbers $a$ and $b$ is 10.

Geometric Mean (G.M.) of two positive numbers $a$ and $b$ is 8.

To Find:

The two positive numbers $a$ and $b$.

Solution:

Let the two positive numbers be $a$ and $b$.

The formula for the Arithmetic Mean of $a$ and $b$ is A.M. $= \frac{a+b}{2}$.

Given A.M. $= 10$, we have:

From equation (i), we can express the sum of the numbers:

The formula for the Geometric Mean of $a$ and $b$ is G.M. $= \sqrt{ab}$.

Given G.M. $= 8$, we have:

Squaring both sides of equation (iii), we get the product of the numbers:

$(\sqrt{ab})^2 = 8^2$

Now we have a system of two equations with two variables:

$a + b = 20$ (from ii)

$ab = 64$ (from iv)

From equation (ii), we can express $b$ in terms of $a$: $b = 20 - a$.

Substitute this expression for $b$ into equation (iv):

$a(20 - a) = 64$

$20a - a^2 = 64$

Rearrange the terms to form a quadratic equation:

$a^2 - 20a + 64 = 0$

We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's try factoring.

We need two numbers that multiply to 64 and add up to -20. These numbers are -16 and -4.

So, we can factor the quadratic equation as:

$(a - 16)(a - 4) = 0$

This equation holds true if either factor is zero.

Case 1: $a - 16 = 0 \implies a = 16$

Case 2: $a - 4 = 0 \implies a = 4$

Now we find the corresponding value of $b$ for each value of $a$ using the equation $b = 20 - a$ (from ii).

If $a = 16$, then $b = 20 - 16 = 4$.

If $a = 4$, then $b = 20 - 4 = 16$.

In both cases, the pair of numbers is 4 and 16.

Since the problem asks for "the numbers", and the order does not matter for A.M. and G.M., the two numbers are 4 and 16.

Let's verify the A.M. and G.M. for the numbers 4 and 16:

A.M. $= \frac{4+16}{2} = \frac{20}{2} = 10$ (Matches the given A.M.)

G.M. $= \sqrt{4 \times 16} = \sqrt{64} = 8$ (Matches the given G.M.)

Final Answer:

The two positive numbers are 4 and 16.

Given:

The Geometric Progression (G.P.) is $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots$

To Find:

The 20^th term ($a_{20}$) and the n^th term ($a_n$) of the G.P.

Solution:

In the given G.P., the first term is $a = \frac{5}{2}$.

To find the common ratio ($r$), we divide the second term by the first term:

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

Alternatively, dividing the third term by the second term:

$r = \frac{5/8}{5/4} = \frac{5}{8} \times \frac{4}{5} = \frac{20}{40} = \frac{1}{2}$

So, the common ratio is $r = \frac{1}{2}$.

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

Finding the 20^th term:

Substitute $a = \frac{5}{2}$, $r = \frac{1}{2}$, and $n = 20$ into the formula for the n^th term:

$a_{20} = a r^{20-1}$

$a_{20} = \frac{5}{2} \left(\frac{1}{2}\right)^{19}$

$a_{20} = \frac{5}{2^1} \cdot \frac{1^{19}}{2^{19}}$

$a_{20} = \frac{5}{2^1} \cdot \frac{1}{2^{19}}$

$a_{20} = \frac{5}{2^{1+19}}$

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

Finding the n^th term:

Substitute $a = \frac{5}{2}$ and $r = \frac{1}{2}$ into the formula for the n^th term:

$a_n = ar^{n-1}$

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

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

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

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

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

Final Answer:

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

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

Given:

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

The common ratio of the G.P., $r = 2$.

To Find:

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

Solution:

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

We are given the 8^th term ($n=8$) and the common ratio ($r=2$). Using the formula for the 8^th term:

$a_8 = ar^{8-1}$

$a_8 = ar^7$

Substitute the given values $a_8 = 192$ and $r = 2$ into the equation:

$192 = a(2)^7$

Calculate $2^7$:

$2^7 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 128$

So, the equation becomes:

$192 = a \times 128$

Now, solve for the first term $a$:

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

Simplify the fraction $\frac{192}{128}$:

Divide both numerator and denominator by their greatest common divisor. Both are divisible by 64.

$192 = 3 \times 64$

$128 = 2 \times 64$

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

So, the first term of the G.P. is $a = \frac{3}{2}$.

Now we need to find the 12^th term ($n=12$). We use the formula $a_n = ar^{n-1}$ again with $a = \frac{3}{2}$, $r = 2$, and $n = 12$:

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

$a_{12} = ar^{11}$

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

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

We have $2^{11} = 2 \times 2^{10} = 2 \times 1024 = 2048$.

$a_{12} = \frac{3}{2} \times 2048$

Cancel out the common factor of 2:

$a_{12} = 3 \times \frac{\cancel{2048}^{1024}}{\cancel{2}_{1}}$

$a_{12} = 3 \times 1024$

Perform the multiplication:

$a_{12} = 3072$

Final Answer:

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

Given:

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

8^th term of a G.P., $a_8 = q$

11^th term of a G.P., $a_{11} = s$

To Show:

$q^2 = ps$

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 $a_n = ar^{n-1}$.

Using this formula, we can write the given terms as:

Now, we need to show that $q^2 = ps$.

Consider the left-hand side (LHS) of the equation, $q^2$. Substitute the expression for $q$ from equation (ii):

LHS $= q^2 = (ar^7)^2 = a^2 (r^7)^2 = a^2 r^{7 \times 2} = a^2 r^{14}$

Consider the right-hand side (RHS) of the equation, $ps$. Substitute the expressions for $p$ from equation (i) and $s$ from equation (iii):

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

Using the properties of exponents ($x^m \cdot x^n = x^{m+n}$ and $(xy)^m = x^m y^m$):

RHS $= a \cdot a \cdot r^4 \cdot r^{10} = a^{1+1} r^{4+10} = a^2 r^{14}$

Comparing the LHS and RHS:

LHS $= a^2 r^{14}$

RHS $= a^2 r^{14}$

Since LHS = RHS, we have shown that $q^2 = ps$.

Conclusion:

The relationship $q^2 = ps$ is true for the 5^th, 8^th, and 11^th terms of the G.P. being $p$, $q$, and $s$ respectively.

Given:

The 4^th term of a G.P. is the square of its second term: $a_4 = (a_2)^2$.

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

To Find:

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

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 second term ($n=2$) is:

The fourth term ($n=4$) is:

We are given that $a_4 = (a_2)^2$. Substitute the expressions from (i) and (ii):

$ar^3 = (ar)^2$

$ar^3 = a^2 r^2$

We are given that the first term $a = -3$. Substitute this value into the equation:

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

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

Move all terms to one side to solve for $r$:

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

Factor out the common term, which is $-3r^2$:

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

For this equation to be true, either $-3r^2 = 0$ or $r+3 = 0$.

Case 1: $-3r^2 = 0 \implies r^2 = 0 \implies r = 0$. However, a common ratio of 0 is generally excluded for a standard G.P. with a non-zero first term, as it makes subsequent terms zero and the ratio undefined after the first term.

Case 2: $r + 3 = 0 \implies r = -3$. This is a valid common ratio.

So, the common ratio is $r = -3$ and the first term is $a = -3$.

Now, we need to find the 7^th term ($n=7$). Use the formula $a_n = ar^{n-1}$ with $a = -3$, $r = -3$, and $n = 7$:

$a_7 = ar^{7-1}$

$a_7 = ar^6$

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

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

Calculate $(-3)^6$. Since the power is even, the result is positive: $(-3)^6 = 3^6 = 729$.

$a_7 = (-3)(729)$

Perform the multiplication:

$a_7 = -2187$

Final Answer:

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

Given:

(a) G.P.: 2, $2\sqrt{2}$, 4, ...; Term value: 128

(b) G.P.: $\sqrt{3}$, 3, $3\sqrt{3}$, ...; Term value: 729

(c) G.P.: $\frac{1}{3}$, $\frac{1}{9}$, $\frac{1}{27}$, ...; Term value: $\frac{1}{19683}$

To Find:

The term number ($n$) for the given term value in each sequence.

Solution (a):

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

The first term is $a = 2$.

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

We need to find which term is 128. Let the n^th term ($a_n$) be 128.

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

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

Divide both sides by 2:

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

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

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

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

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

Equating the exponents since the bases are the same:

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

Multiply both sides by 2:

$12 = n - 1$

Add 1 to both sides:

$n = 12 + 1$

$n = 13$

Solution (b):

The given G.P. is $\sqrt{3}$, 3, $3\sqrt{3}$, ...

The first term is $a = \sqrt{3}$.

The common ratio is $r = \frac{3}{\sqrt{3}} = \sqrt{3}$.

We need to find which term is 729. Let the n^th term ($a_n$) be 729.

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

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

Using the property of exponents $x^m \cdot x^n = x^{m+n}$:

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

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

Express both sides with the same base. We know that $\sqrt{3} = 3^{1/2}$ and $729 = 3^6$ ($3^2=9$, $3^3=27$, $3^4=81$, $3^5=243$, $3^6=729$).

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

$3^6 = 3^{n/2}$

Equating the exponents since the bases are the same:

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

Multiply both sides by 2:

$n = 12$

Solution (c):

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

The first term is $a = \frac{1}{3}$.

The common ratio is $r = \frac{1/9}{1/3} = \frac{1}{9} \times \frac{3}{1} = \frac{3}{9} = \frac{1}{3}$.

We need to find which term is $\frac{1}{19683}$. Let the n^th term ($a_n$) be $\frac{1}{19683}$.

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

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

Using the property of exponents $x^m \cdot x^n = x^{m+n}$:

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

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

This can also be written as:

$19683 = 3^n$

We need to find the power of 3 that equals 19683.

$3^1=3$, $3^2=9$, $3^3=27$, $3^4=81$, $3^5=243$, $3^6=729$, $3^7=2187$, $3^8=6561$, $3^9=19683$.

So, $3^n = 3^9$.

Equating the exponents:

$n = 9$

Final Answer:

(a) The 13^th term of the G.P. 2, $2\sqrt{2}$, 4, ... is 128.

(b) The 12^th term of the G.P. $\sqrt{3}$, 3, $3\sqrt{3}$, ... is 729.

(c) The 9^th term of the G.P. $\frac{1}{3}$, $\frac{1}{9}$, $\frac{1}{27}$, ... is $\frac{1}{19683}$.

Given:

The three numbers are $-\frac{2}{7}$, x, $-\frac{7}{2}$.

These numbers are in G.P.

To Find:

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

Solution:

For three numbers $a, b, c$ to be in Geometric Progression (G.P.), the ratio of consecutive terms must be constant. This means $\frac{b}{a} = \frac{c}{b}$.

In this case, the numbers are $a_1 = -\frac{2}{7}$, $a_2 = x$, and $a_3 = -\frac{7}{2}$.

Applying the condition for a G.P.:

$\frac{a_2}{a_1} = \frac{a_3}{a_2}$

Substitute the given terms:

$\frac{x}{-2/7} = \frac{-7/2}{x}$

To solve for $x$, we can cross-multiply:

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

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

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

$x^2 = 1$

To find the value(s) of $x$, take the square root of both sides:

$x = \pm \sqrt{1}$

$x = \pm 1$

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

Let's verify the sequences for both values of $x$:

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

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

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

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

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

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

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

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

Final Answer:

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

Given:

The Geometric Progression (G.P.) is 0.15, 0.015, 0.0015, ...

Number of terms to sum is $n = 20$.

To Find:

The sum of the first 20 terms of the G.P. ($S_{20}$).

Solution:

In the given G.P., the first term is $a = 0.15$.

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

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

So, the common ratio is $r = 0.1 = \frac{1}{10}$.

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

Since $r = 0.1$, we have $|r| = 0.1 < 1$, so we can use this formula.

Substitute $a = 0.15$, $r = 0.1$, and $n = 20$ into the formula:

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

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

We can simplify the fraction $\frac{0.15}{0.9}$:

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

So, the sum is:

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

Since $(0.1)^{20} = (\frac{1}{10})^{20} = \frac{1}{10^{20}}$, we can write:

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

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

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

Final Answer:

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

Given:

The Geometric Progression (G.P.) is $\sqrt{7}, \sqrt{21}, 3\sqrt{7}, \dots$

The sum is required for $n$ terms.

To Find:

The sum of the first $n$ terms of the G.P. ($S_n$).

Solution:

In the given G.P., the first term is $a = \sqrt{7}$.

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

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

Alternatively, using the third term and the second term:

$r = \frac{3\sqrt{7}}{\sqrt{21}} = \frac{3\sqrt{7}}{\sqrt{3} \sqrt{7}} = \frac{3}{\sqrt{3}} = \frac{3\sqrt{3}}{(\sqrt{3})^2} = \frac{3\sqrt{3}}{3} = \sqrt{3}$

So, the common ratio is $r = \sqrt{3}$.

The number of terms is $n$.

The sum of the first $n$ terms of a G.P. is given by the formula $S_n = \frac{a(r^n - 1)}{r - 1}$, when $|r| \neq 1$.

Since $r = \sqrt{3}$ (which is approximately 1.732), $|r| > 1$, so we use this formula.

Substitute $a = \sqrt{7}$ and $r = \sqrt{3}$ into the formula:

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

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

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

It is common practice to rationalize the denominator. Multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} + 1$:

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

Using the difference of squares formula in the denominator, $(x-y)(x+y) = x^2 - y^2$:

Denominator $= (\sqrt{3})^2 - 1^2 = 3 - 1 = 2$

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

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

We can also write $\sqrt{7}(\sqrt{3} + 1)$ as $\sqrt{21} + \sqrt{7}$.

$S_n = \frac{(\sqrt{21} + \sqrt{7})(3^{n/2} - 1)}{2}$

Final Answer:

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

Given:

The Geometric Progression (G.P.) is $1, -a, a^2, -a^3, \dots$

Number of terms to sum is $n$.

Constraint: $a \neq -1$.

To Find:

The sum of the first $n$ terms of the G.P. ($S_n$).

Solution:

In the given G.P., the first term is $A = 1$.

The common ratio ($R$) is found by dividing a term by its preceding term:

$R = \frac{-a}{1} = -a$

The common ratio is $R = -a$.

We are given that $a \neq -1$, which implies $-a \neq -(-1)$, so $-a \neq 1$.

Thus, the common ratio $R \neq 1$.

The sum of the first $n$ terms of a G.P. with first term $A$ and common ratio $R \neq 1$ is given by the formula:

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

Substitute the values $A = 1$ and $R = -a$ into the formula:

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

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

We can move the negative sign from the denominator to the numerator:

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

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

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

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

The term $(-a)^n$ can be written as $(-1)^n a^n$.

So, the sum can also be written as:

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

This formula is valid for any positive integer $n$ and any value of $a \neq -1$.

Final Answer:

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

Given:

The Geometric Progression (G.P.) is $x^3, x^5, x^7, \dots$

Number of terms to sum is $n$.

Constraint: $x \neq \pm 1$.

To Find:

The sum of the first $n$ terms of the G.P. ($S_n$).

Solution:

In the given G.P., the first term is $a = x^3$.

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

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

Alternatively, using the third term and the second term:

$r = \frac{x^7}{x^5} = x^{7-5} = x^2$

So, the common ratio is $r = x^2$.

We are given that $x \neq \pm 1$.

If $x = 1$, then $r = 1^2 = 1$. If $x = -1$, then $r = (-1)^2 = 1$.

Since the problem states $x \neq \pm 1$, the common ratio $r = x^2$ is not equal to 1.

The sum of the first $n$ terms of a G.P. with first term $a$ and common ratio $r \neq 1$ is given by the formula:

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

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

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

Using the property of exponents $(y^m)^n = y^{mn}$:

$(x^2)^n = x^{2n}$

So, the sum is:

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

This formula is valid for any positive integer $n$ and any value of $x \neq \pm 1$.

Final Answer:

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

Given:

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

To Evaluate:

The sum of the given expression from $k=1$ to $k=11$.

Solution:

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

We can split the summation into two parts:

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

First, evaluate the sum of the constant term $\sum\limits_{k=1}^{11} 2$. This is the sum of 2 added 11 times.

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

Next, evaluate the sum of the second term $\sum\limits_{k=1}^{11} 3^k$. This represents the sum of the sequence $3^1, 3^2, 3^3, \dots, 3^{11}$.

The sequence is $3, 9, 27, \dots, 3^{11}$.

This is a Geometric Progression (G.P.) with:

First term, $a = 3$

Common ratio, $r = \frac{9}{3} = 3$

Number of terms, $n = 11$

Since the common ratio $r=3$ is not equal to 1, the sum of the first $n$ terms of a G.P. is given by the formula $S_n = \frac{a(r^n - 1)}{r - 1}$.

Substitute the values $a=3$, $r=3$, and $n=11$ into the formula:

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

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

Now, we calculate $3^{11}$.

$3^{11} = 177147$

Substitute the value of $3^{11}$ into the sum of the G.P.:

$\sum\limits_{k=1}^{11} 3^k = \frac{3(177147 - 1)}{2}$

$\sum\limits_{k=1}^{11} 3^k = \frac{3(177146)}{2}$

$\sum\limits_{k=1}^{11} 3^k = 3 \times 88573$

$\sum\limits_{k=1}^{11} 3^k = 265719$

Finally, add the results of the two parts:

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

$\sum\limits_{k=1}^{11} (2 + 3^k) = 22 + 265719$

$\sum\limits_{k=1}^{11} (2 + 3^k) = 265741$

Final Answer:

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

Given:

Sum of the first three terms of a G.P. $= \frac{39}{10}$

Product of the first three terms of a G.P. $= 1$

To Find:

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

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.

According to the problem, the product of the first three terms is 1.

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

$a^3 = 1$

Taking the cube root of both sides, we get:

According to the problem, the sum of the first three terms is $\frac{39}{10}$.

$\frac{a}{r} + a + ar = \frac{39}{10}$

Substitute the value of $a = 1$ from (i) into this equation:

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

Multiply the entire equation by $10r$ to eliminate the fractions (assuming $r \neq 0$, which must be true for a G.P.):

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

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

Move all terms to one side to get a standard quadratic equation:

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

$10r^2 - 29r + 10 = 0$

This is a quadratic equation of the form $Ar^2 + Br + C = 0$, where $A=10$, $B=-29$, and $C=10$. We can solve for $r$ using the quadratic formula $r = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$.

$r = \frac{-(-29) \pm \sqrt{(-29)^2 - 4(10)(10)}}{2(10)}$

$r = \frac{29 \pm \sqrt{841 - 400}}{20}$

$r = \frac{29 \pm \sqrt{441}}{20}$

We know that $\sqrt{441} = 21$.

$r = \frac{29 \pm 21}{20}$

We have two possible values for the common ratio $r$:

Case 1: $r = \frac{29 + 21}{20} = \frac{50}{20} = \frac{5}{2}$

Case 2: $r = \frac{29 - 21}{20} = \frac{8}{20} = \frac{2}{5}$

Now we find the terms of the G.P. for each case, using $a=1$ from (i).

Case 1: $a = 1$ and $r = \frac{5}{2}$

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

Second term = $a = 1$

Third term = $ar = (1)(\frac{5}{2}) = \frac{5}{2}$

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

Case 2: $a = 1$ and $r = \frac{2}{5}$

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

Second term = $a = 1$

Third term = $ar = (1)(\frac{2}{5}) = \frac{2}{5}$

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

The set of terms obtained in both cases is the same, $\{\frac{2}{5}, 1, \frac{5}{2}\}$, just in a different order depending on which value of $r$ is considered the common ratio from left to right.

Therefore, the common ratio is either $\frac{5}{2}$ or $\frac{2}{5}$, and the terms of the G.P. are $\frac{2}{5}, 1,$ and $\frac{5}{2}$.

Final Answer:

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

The terms of the G.P. are $\frac{2}{5}, 1, \frac{5}{2}$.

Given:

The Geometric Progression (G.P.) is 3, $3^2$, $3^3$, ...

The sum of the G.P. is $S_n = 120$.

To Find:

The number of terms ($n$) required to obtain the sum 120.

Solution:

The given G.P. is 3, 9, 27, ...

The first term is $a = 3$.

The common ratio is $r = \frac{9}{3} = 3$.

The sum of the first $n$ terms of a G.P. is given by the formula $S_n = \frac{a(r^n - 1)}{r - 1}$, since the common ratio $r=3$ is not equal to 1.

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

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

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

Multiply both sides of the equation 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:

$80 + 1 = 3^n$

$81 = 3^n$

To find the value of $n$, we need to express 81 as a power of 3.

We know that $3^4 = 3 \times 3 \times 3 \times 3 = 9 \times 9 = 81$.

So, $81 = 3^4$.

The equation becomes:

$3^4 = 3^n$

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

$n = 4$

Final Answer:

4 terms of the G.P. 3, $3^2$, $3^3$, ... are needed to give the sum 120.

Given:

Sum of the first three terms of a G.P. ($S_3$) $= 16$.

Sum of the next three terms (4^th, 5^th, and 6^th terms) $= 128$.

To Find:

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

Solution:

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

The terms of the G.P. are $a, ar, ar^2, ar^3, ar^4, ar^5, \dots$

The sum of the first three terms is:

$a + ar + ar^2 = 16$

Factor out $a$ from the left side:

The next three terms are the 4^th, 5^th, and 6^th terms, which are $ar^3, ar^4, ar^5$.

The sum of the next three terms is:

$ar^3 + ar^4 + ar^5 = 128$

Factor out $ar^3$ from the left side:

Now we have a system of two equations:

1) $a(1 + r + r^2) = 16$

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

Divide equation (ii) by equation (i). We can do this assuming $a \neq 0$ (otherwise the terms and sums would be 0) and $1 + r + r^2 \neq 0$. Since $1+r+r^2 = (r + 1/2)^2 + 3/4$, it is always positive for real $r$, so $1+r+r^2 \neq 0$.

$\frac{ar^3(1 + r + r^2)}{a(1 + r + r^2)} = \frac{128}{16}$

Cancel the common terms $a$ and $(1 + r + r^2)$:

$r^3 = \frac{128}{16}$

$r^3 = 8$

Take the cube root of both sides to find $r$:

$r = \sqrt[3]{8}$

$r = 2$

Now substitute the value of $r = 2$ into equation (i) to find the first term $a$:

$a(1 + r + r^2) = 16$

$a(1 + 2 + 2^2) = 16$

$a(1 + 2 + 4) = 16$

$a(7) = 16$

Divide by 7 to solve for $a$:

$a = \frac{16}{7}$

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

We have $a = \frac{16}{7}$ and $r = 2$. Substitute these values into the sum formula:

$S_n = \frac{\frac{16}{7}(2^n - 1)}{2 - 1}$

$S_n = \frac{\frac{16}{7}(2^n - 1)}{1}$

$S_n = \frac{16}{7}(2^n - 1)$

Final 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 a G.P., $a = 729$.

The 7^th term of the G.P., $a_7 = 64$.

To Find:

The sum of the first 7 terms of the G.P. ($S_7$).

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 $a_n = ar^{n-1}$.

For the 7^th term ($n=7$), the formula is:

$a_7 = ar^{7-1} = ar^6$

Substitute the given values $a = 729$ and $a_7 = 64$:

$64 = 729 \cdot r^6$

Divide both sides by 729:

$r^6 = \frac{64}{729}$

To find $r$, take the sixth root of both sides:

$r = \sqrt[6]{\frac{64}{729}}$

We know that $64 = 2^6$ and $729 = 3^6$.

$r^6 = \left(\frac{2}{3}\right)^6$

This gives two possible real values for $r$: $r = \frac{2}{3}$ or $r = -\frac{2}{3}$.

Since both the first term (729) and the 7th term (64) are positive, a G.P. with terms $a, ar, ar^2, \dots$ would have terms of the same sign if $r$ is positive. If $r$ is negative, the terms would alternate in sign ($+, -, +, -, \dots, +$). Both cases are mathematically valid based on the given terms $a_1$ and $a_7$. We will calculate the sum $S_7$ for the common ratio $r = \frac{2}{3}$ first, as sequences with terms of consistent sign are often implied unless stated otherwise.

Case 1: Common ratio $r = \frac{2}{3}$

We have $a = 729$, $r = \frac{2}{3}$, and $n = 7$.

The common ratio $|r| = |\frac{2}{3}| < 1$, so the formula for the sum of the first $n$ terms is $S_n = \frac{a(1-r^n)}{1-r}$.

Substitute the values for $S_7$:

$S_7 = \frac{729 \left(1 - \left(\frac{2}{3}\right)^7\right)}{1 - \frac{2}{3}}$

Calculate $\left(\frac{2}{3}\right)^7$:

$\left(\frac{2}{3}\right)^7 = \frac{2^7}{3^7} = \frac{128}{2187}$

Calculate the denominator $1 - \frac{2}{3}$:

$1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

Substitute these values back into the formula for $S_7$:

$S_7 = \frac{729 \left(1 - \frac{128}{2187}\right)}{\frac{1}{3}}$

Simplify the term in the parenthesis:

$1 - \frac{128}{2187} = \frac{2187}{2187} - \frac{128}{2187} = \frac{2187 - 128}{2187} = \frac{2059}{2187}$

Substitute this back:

$S_7 = \frac{729 \left(\frac{2059}{2187}\right)}{\frac{1}{3}}$

$S_7 = 729 \times \frac{2059}{2187} \times 3$

Note that $2187 = 3^7$ and $729 = 3^6$. So, $\frac{729}{2187} = \frac{3^6}{3^7} = \frac{1}{3}$.

$S_7 = \cancel{729}^{1} \times \frac{2059}{\cancel{2187}_{3}} \times \cancel{3}^{1}$

$S_7 = 1 \times 2059 \times 1 = 2059$

Alternate Solution (Case 2: Common ratio $r = -\frac{2}{3}$):

If the common ratio is $r = -\frac{2}{3}$, we still have $a = 729$ and $n = 7$.

The common ratio $|r| = |-\frac{2}{3}| = \frac{2}{3} < 1$, so the formula for the sum is still $S_n = \frac{a(1-r^n)}{1-r}$.

Substitute the values for $S_7$:

$S_7 = \frac{729 \left(1 - \left(-\frac{2}{3}\right)^7\right)}{1 - \left(-\frac{2}{3}\right)}$

Calculate $\left(-\frac{2}{3}\right)^7$:

$\left(-\frac{2}{3}\right)^7 = \frac{(-2)^7}{3^7} = \frac{-128}{2187}$

Calculate the denominator $1 - \left(-\frac{2}{3}\right)$:

$1 - \left(-\frac{2}{3}\right) = 1 + \frac{2}{3} = \frac{3}{3} + \frac{2}{3} = \frac{5}{3}$

Substitute these values back into the formula for $S_7$:

$S_7 = \frac{729 \left(1 - \left(-\frac{128}{2187}\right)\right)}{\frac{5}{3}}$

$S_7 = \frac{729 \left(1 + \frac{128}{2187}\right)}{\frac{5}{3}}$

Simplify the term in the parenthesis:

$1 + \frac{128}{2187} = \frac{2187}{2187} + \frac{128}{2187} = \frac{2187 + 128}{2187} = \frac{2315}{2187}$

Substitute this back:

$S_7 = \frac{729 \left(\frac{2315}{2187}\right)}{\frac{5}{3}}$

$S_7 = 729 \times \frac{2315}{2187} \times \frac{3}{5}$

Using $\frac{729}{2187} = \frac{1}{3}$ as calculated before:

$S_7 = \cancel{729}^{1} \times \frac{2315}{\cancel{2187}_{3}} \times \frac{\cancel{3}^{1}}{5}$

$S_7 = \frac{2315}{5}$

$S_7 = 463$

Both 2059 (for $r=2/3$) and 463 (for $r=-2/3$) are valid sums depending on the specific G.P. intended. Without further information (e.g., if terms are positive, or if the sequence follows a specific pattern implied by the notation), both are possible.

Final Answer:

Based on the context usually implying a positive common ratio when terms are positive, the sum of the first 7 terms is 2059.

However, if the common ratio is negative, the sum is 463.

Let the Geometric Progression (G.P.) be denoted by $a, ar, ar^2, ar^3, \dots$.

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

The $n$-th term of a G.P. is given by $a_n = ar^{n-1}$.

Given:

The sum of the first two terms is $-4$.

The fifth term is 4 times the third term.

To Find:

A G.P. satisfying the given conditions.

Solution:

According to the first condition, the sum of the first two terms ($a_1$ and $a_2$) is $-4$.

$a_1 + a_2 = -4$

Factor out $a$ from the left side of equation (i):

According to the second condition, the fifth term ($a_5$) is 4 times the third term ($a_3$).

$a_5 = 4 \times a_3$

Now, we solve equation (ii) for $r$.

$ar^4 - 4ar^2 = 0$

Factor out $ar^2$:

This equation holds if $a=0$ or $r^2=0$ or $r^2-4=0$.

If $a=0$, then from equation (i), $0(1+r) = -4$, which means $0 = -4$, which is false. So, $a$ cannot be $0$.

Since $a \neq 0$, we must have $r^2(r^2 - 4) = 0$.

This implies either $r^2 = 0$ or $r^2 - 4 = 0$.

Case 1: $r^2 = 0$

This gives $r = 0$.

Substitute $r=0$ into equation (i), $a(1 + r) = -4$:

$a(1 + 0) = -4$

$a(1) = -4$

$a = -4$

In this case, the first term is $a = -4$ and the common ratio is $r = 0$.

The G.P. is $-4, -4(0), -4(0)^2, -4(0)^3, \dots$, which is $-4, 0, 0, 0, \dots$.

Case 2: $r^2 - 4 = 0$

This gives $r^2 = 4$, so $r = \sqrt{4}$ or $r = -\sqrt{4}$.

$r = 2$ or $r = -2$.

Subcase 2a: $r = 2$

Substitute $r=2$ into equation (i), $a(1 + r) = -4$:

$a(1 + 2) = -4$

$3a = -4$

$a = \frac{-4}{3}$

In this case, the first term is $a = \frac{-4}{3}$ and the common ratio is $r = 2$.

The G.P. is $\frac{-4}{3}, \left(\frac{-4}{3}\right)(2), \left(\frac{-4}{3}\right)(2)^2, \left(\frac{-4}{3}\right)(2)^3, \dots$, which is $\frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}, \frac{-32}{3}, \dots$.

Subcase 2b: $r = -2$

Substitute $r=-2$ into equation (i), $a(1 + r) = -4$:

$a(1 + (-2)) = -4$

$a(-1) = -4$

$-a = -4$

$a = 4$

In this case, the first term is $a = 4$ and the common ratio is $r = -2$.

The G.P. is $4, 4(-2), 4(-2)^2, 4(-2)^3, \dots$, which is $4, -8, 16, -32, \dots$.

Thus, there are three possible Geometric Progressions that satisfy the given conditions:

1. First term $-4$, Common ratio $0$. The G.P. is $-4, 0, 0, 0, \dots$.

2. First term $\frac{-4}{3}$, Common ratio $2$. The G.P. is $\frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}, \frac{-32}{3}, \dots$.

3. First term $4$, Common ratio $-2$. The G.P. is $4, -8, 16, -32, \dots$.

Given:

The 4^th term of a G.P. is $x$.

The 10^th term of a G.P. is $y$.

The 16^th term of a G.P. is $z$.

To Prove:

x, y, and z are in G.P.

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 $a_n = AR^{n-1}$.

Using this formula, we can write the given terms:

For three numbers x, y, and z to be in G.P., the square of the middle term must be equal to the product of the other two terms, i.e., $y^2 = xz$.

Consider the left-hand side (LHS) of the equation, $y^2$. Substitute the expression for $y$ from equation (ii):

LHS $= y^2 = (AR^9)^2 = A^2 (R^9)^2 = A^2 R^{9 \times 2} = A^2 R^{18}$

Consider the right-hand side (RHS) of the equation, $xz$. Substitute the expressions for $x$ from equation (i) and $z$ from equation (iii):

RHS $= xz = (AR^3)(AR^{15})$

Using the properties of exponents $(st)^m = s^m t^m$ and $s^m \cdot s^n = s^{m+n}$:

RHS $= A \cdot A \cdot R^3 \cdot R^{15} = A^{1+1} R^{3+15} = A^2 R^{18}$

Comparing the LHS and RHS:

LHS $= A^2 R^{18}$

RHS $= A^2 R^{18}$

Since LHS = RHS, we have $y^2 = xz$.

This condition proves that x, y, and z are in G.P.

Conclusion:

Therefore, the 4^th, 10^th, and 16^th terms of a G.P., which are x, y, and z respectively, are themselves in G.P.

Given:

The sequence is 8, 88, 888, 8888, ...

To Find:

The sum of the sequence to $n$ terms.

Solution:

Let the sum of the sequence to $n$ terms be $S_n$.

$S_n = 8 + 88 + 888 + \dots + \underbrace{88\dots8}_{n \text{ terms}}$

We can write each term as a product of 8 and a sequence of 1s.

$S_n = 8(1) + 8(11) + 8(111) + \dots + 8(\underbrace{11\dots1}_{n \text{ terms}})$

Factor out 8:

$S_n = 8 (1 + 11 + 111 + \dots + \underbrace{11\dots1}_{n \text{ terms}})$

Now, we can express the terms within the parenthesis using powers of 10. Note that $\underbrace{11\dots1}_{k \text{ terms}} = \frac{10^k - 1}{9}$.

So, $1 = \frac{10^1 - 1}{9}$, $11 = \frac{10^2 - 1}{9}$, $111 = \frac{10^3 - 1}{9}$, and so on.

$S_n = 8 \left( \frac{10^1 - 1}{9} + \frac{10^2 - 1}{9} + \frac{10^3 - 1}{9} + \dots + \frac{10^n - 1}{9} \right)$

Factor out $\frac{1}{9}$:

$S_n = \frac{8}{9} \left[ (10^1 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1) \right]$

Group the terms involving powers of 10 and the $-1$ terms separately:

$S_n = \frac{8}{9} \left[ (10^1 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + 1 \text{ ($n$ times)}) \right]$

$S_n = \frac{8}{9} \left[ (10^1 + 10^2 + 10^3 + \dots + 10^n) - n \right]$

The expression $10^1 + 10^2 + 10^3 + \dots + 10^n$ is a geometric series with first term $a = 10$, common ratio $r = 10$, and $n$ terms. The sum of this G.P. is given by the formula $S_{\text{G.P.}} = \frac{a(r^n - 1)}{r - 1}$.

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

Substitute this back into the expression for $S_n$:

$S_n = \frac{8}{9} \left[ \frac{10(10^n - 1)}{9} - n \right]$

To simplify further, we can find a common denominator inside the bracket:

$S_n = \frac{8}{9} \left[ \frac{10(10^n - 1)}{9} - \frac{9n}{9} \right]$

$S_n = \frac{8}{9} \left[ \frac{10(10^n - 1) - 9n}{9} \right]$

$S_n = \frac{8}{81} [10(10^n - 1) - 9n]$

$S_n = \frac{8}{81} [10^{n+1} - 10 - 9n]$

Final Answer:

The sum of the sequence 8, 88, 888, ... to $n$ terms is $S_n = \frac{8}{81} (10^{n+1} - 9n - 10)$.

Given:

The first sequence is: 2, 4, 8, 16, 32

The second sequence is: 128, 32, 8, 2, $\frac{1}{2}$

To Find:

The sum of the products of the corresponding terms of the two sequences.

Solution:

Let the first sequence be denoted by $\{a_n\}$ and the second sequence by $\{b_n\}$.

The terms of the first sequence are $a_1=2, a_2=4, a_3=8, a_4=16, a_5=32$. This is a G.P. with first term $a=2$ and common ratio $r_a = \frac{4}{2} = 2$.

The terms of the second sequence are $b_1=128, b_2=32, b_3=8, b_4=2, b_5=\frac{1}{2}$. This is a G.P. with first term $b=128$ and common ratio $r_b = \frac{32}{128} = \frac{1}{4}$.

We need to find the sum of the products of the corresponding terms: $c_n = a_n \times b_n$ for $n=1, 2, 3, 4, 5$. The sum is $S = \sum_{n=1}^{5} c_n = c_1 + c_2 + c_3 + c_4 + c_5$.

Let's find the terms of the product sequence $\{c_n\}$:

$c_1 = a_1 \times b_1 = 2 \times 128 = 256$

$c_2 = a_2 \times b_2 = 4 \times 32 = 128$

$c_3 = a_3 \times b_3 = 8 \times 8 = 64$

$c_4 = a_4 \times b_4 = 16 \times 2 = 32$

$c_5 = a_5 \times b_5 = 32 \times \frac{1}{2} = 16$

The sequence of products is 256, 128, 64, 32, 16.

Let's check if this new sequence is a G.P. We examine the ratio of consecutive terms:

Ratio 1: $\frac{c_2}{c_1} = \frac{128}{256} = \frac{1}{2}$

Ratio 2: $\frac{c_3}{c_2} = \frac{64}{128} = \frac{1}{2}$

Ratio 3: $\frac{c_4}{c_3} = \frac{32}{64} = \frac{1}{2}$

Ratio 4: $\frac{c_5}{c_4} = \frac{16}{32} = \frac{1}{2}$

Since the ratio of consecutive terms is constant ($\frac{1}{2}$), the sequence of products $\{c_n\}$ is a G.P. with first term $A = 256$ and common ratio $R = \frac{1}{2}$. Note that the common ratio of the product G.P. is $R = r_a \times r_b = 2 \times \frac{1}{4} = \frac{1}{2}$.

We need to find the sum of these 5 terms ($n=5$). The sum of the first $n$ terms of a G.P. with first term $A$ and common ratio $R$ ($|R| < 1$) is $S_n = \frac{A(1-R^n)}{1-R}$.

Here, $A = 256$, $R = \frac{1}{2}$, and $n = 5$.

$S_5 = \frac{256 \left(1 - \left(\frac{1}{2}\right)^5\right)}{1 - \frac{1}{2}}$

$S_5 = \frac{256 \left(1 - \frac{1}{32}\right)}{\frac{1}{2}}$

$S_5 = \frac{256 \left(\frac{32-1}{32}\right)}{\frac{1}{2}}$

$S_5 = \frac{256 \left(\frac{31}{32}\right)}{\frac{1}{2}}$

$S_5 = 256 \times \frac{31}{32} \times 2$

$S_5 = \cancel{256}^{8} \times \frac{31}{\cancel{32}_{1}} \times 2$

$S_5 = 8 \times 31 \times 2$

$S_5 = 16 \times 31$

$S_5 = 496$

Final Answer:

The sum of the products of the corresponding terms is 496.

Given:

Two sequences which are G.P.s:

Sequence 1: $a, ar, ar^2, \dots, ar^{n-1}$

Sequence 2: $A, AR, AR^2, \dots, AR^{n-1}$

To Prove:

The sequence formed by the products of the corresponding terms is a G.P., and to find its common ratio.

Solution:

Let the first sequence be denoted by $\{a_k\}$ where $a_k = ar^{k-1}$ for $k=1, 2, \dots, n$. The first term is $a$ and the common ratio is $r$.

Let the second sequence be denoted by $\{b_k\}$ where $b_k = AR^{k-1}$ for $k=1, 2, \dots, n$. The first term is $A$ and the common ratio is $R$.

Consider a new sequence $\{c_k\}$ formed by the product of the corresponding terms of the two sequences.

The $k$-th term of this new sequence is $c_k = a_k \times b_k$.

$c_k = (ar^{k-1}) \times (AR^{k-1})$

$c_k = (aA)(r^{k-1}R^{k-1})$

Using the property of exponents $(xy)^m = x^m y^m$, we can write $r^{k-1}R^{k-1} = (rR)^{k-1}$.

So, the $k$-th term of the product sequence is $c_k = (aA)(rR)^{k-1}$.

The terms of the product sequence are:

$c_1 = (aA)(rR)^{1-1} = aA(rR)^0 = aA$

$c_2 = (aA)(rR)^{2-1} = aA(rR)^1 = aARr$

$c_3 = (aA)(rR)^{3-1} = aA(rR)^2 = aA r^2 R^2$

and so on, up to $c_n = (aA)(rR)^{n-1}$.

To show that the sequence $\{c_k\}$ is a G.P., we need to verify that the ratio of any term to its preceding term is constant. That is, $\frac{c_{k+1}}{c_k}$ must be a constant value for all valid $k$.

The $(k+1)$-th term is $c_{k+1} = (aA)(rR)^{(k+1)-1} = (aA)(rR)^k$.

Consider the ratio $\frac{c_{k+1}}{c_k}$:

$\frac{c_{k+1}}{c_k} = \frac{(aA)(rR)^k}{(aA)(rR)^{k-1}}$

Cancel out the term $(aA)$, assuming $aA \neq 0$ (if $a=0$ or $A=0$, all terms of the product sequence would be 0, which can be considered a trivial G.P. with common ratio undefined or any real number depending on definition, but for non-zero terms):

$\frac{c_{k+1}}{c_k} = \frac{(rR)^k}{(rR)^{k-1}}$

Using the property of exponents $\frac{x^m}{x^n} = x^{m-n}$:

$\frac{c_{k+1}}{c_k} = (rR)^{k - (k-1)}$

$\frac{c_{k+1}}{c_k} = (rR)^{k - k + 1}$

$\frac{c_{k+1}}{c_k} = (rR)^1 = rR$

The ratio of consecutive terms is $rR$, which is a constant value since $r$ and $R$ are the common ratios of the original G.P.s.

Since the ratio of consecutive terms is constant, the sequence $\{c_k\}$ formed by the products of the corresponding terms is a Geometric Progression.

The first term of this new G.P. is $c_1 = aA$.

The common ratio of this new G.P. is $rR$.

Conclusion:

The sequence formed by the products of the corresponding terms of the given G.P.s is a Geometric Progression with the first term $aA$ and the common ratio $rR$.

Given:

Four numbers form a G.P.

Condition 1: The third term is greater than the first term by 9.

Condition 2: The second term is greater than the 4^th term by 18.

To Find:

The four numbers in the G.P.

Solution:

Let the four terms of the geometric progression be $a, ar, ar^2, ar^3$, where $a$ is the first term and $r$ is the common ratio.

According to the first condition:

Third term = First term + 9

$ar^2 = a + 9$

Rearranging the terms:

$ar^2 - a = 9$

According to the second condition:

Second term = Fourth term + 18

$ar = ar^3 + 18$

Rearranging the terms:

$ar - ar^3 = 18$

Factor out $ar$:

We now have a system of two equations with two variables $a$ and $r$:

(i) $a(r^2 - 1) = 9$

(ii) $ar(1 - r^2) = 18$

Note that $(1 - r^2) = -(r^2 - 1)$. Substitute this into equation (ii):

$ar(-(r^2 - 1)) = 18$

$-ar(r^2 - 1) = 18$

Divide this equation by equation (i). We can do this if $a \neq 0$ and $r^2 - 1 \neq 0$. If $a=0$, the terms are $0,0,0,0$, which does not satisfy condition 1 ($0=0+9$ is false). If $r^2=1$, then $r = \pm 1$. If $r=1$, equation (i) gives $a(1-1)=9 \implies 0=9$, impossible. If $r=-1$, equation (i) gives $a((-1)^2-1)=9 \implies a(0)=9 \implies 0=9$, impossible. Thus $a \neq 0$ and $r \neq \pm 1$, so $r^2 - 1 \neq 0$.

$\frac{-ar(r^2 - 1)}{a(r^2 - 1)} = \frac{18}{9}$

Cancel the common terms $a$ and $(r^2 - 1)$:

$-r = 2$

Multiply by -1:

$r = -2$

Now substitute the value of $r = -2$ into equation (i) to find the first term $a$:

$a(r^2 - 1) = 9$

$a((-2)^2 - 1) = 9$

$a(4 - 1) = 9$

$a(3) = 9$

Divide by 3:

$a = \frac{9}{3}$

$a = 3$

The four numbers in the G.P. are $a, ar, ar^2, ar^3$. Substitute $a=3$ and $r=-2$:

First term: $a = 3$

Second term: $ar = 3(-2) = -6$

Third term: $ar^2 = 3(-2)^2 = 3(4) = 12$

Fourth term: $ar^3 = 3(-2)^3 = 3(-8) = -24$

The four numbers are 3, -6, 12, -24.

Verification:

The numbers 3, -6, 12, -24 form a G.P. with common ratio -2.

Condition 1: Third term (12) and first term (3). $12 = 3 + 9$. This is true.

Condition 2: Second term (-6) and fourth term (-24). $-6 = -24 + 18$. This is true.

The conditions are satisfied.

Final Answer:

The four numbers forming the geometric progression are 3, -6, 12, and -24.

Given:

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

The q^th term of a G.P. is $b$.

The r^th term of a G.P. is $c$.

To Prove:

$a^{q - r} b^{r - p} c^{p - q} = 1$.

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 for the given terms:

$a_p = AR^{p-1} = a$

$a_q = AR^{q-1} = b$

$a_r = AR^{r-1} = c$

Consider the left-hand side (LHS) of the equation we need to prove:

LHS $= a^{q - r} b^{r - p} c^{p - q}$

Substitute the expressions for $a, b,$ and $c$ in terms of $A$ and $R$:

LHS $= (AR^{p-1})^{q - r} (AR^{q-1})^{r - p} (AR^{r-1})^{p - q}$

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

LHS $= A^{q - r} (R^{p-1})^{q - r} \cdot A^{r - p} (R^{q-1})^{r - p} \cdot A^{p - q} (R^{r-1})^{p - q}$

LHS $= A^{q - r} R^{(p-1)(q - r)} \cdot A^{r - p} R^{(q-1)(r - p)} \cdot A^{p - q} R^{(r-1)(p - q)}$

Group the terms with base $A$ and the terms with base $R$:

LHS $= A^{(q - r) + (r - p) + (p - q)} \cdot R^{(p-1)(q - r) + (q-1)(r - p) + (r-1)(p - q)}$

Evaluate the exponent of $A$:

Exponent of $A = (q - r) + (r - p) + (p - q) = q - r + r - p + p - q = 0$

So, the term with $A$ is $A^0$.

Now, evaluate the exponent of $R$. Expand each product in the exponent:

$(p-1)(q - r) = pq - pr - q + r$

$(q-1)(r - p) = qr - qp - r + p$

$(r-1)(p - q) = rp - rq - p + q$

Sum of the exponents of $R = (pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q)$

Sum of the exponents of $R = pq - pr - q + r + qr - qp - r + p + rp - rq - p + q$

Group like terms:

Sum of the exponents of $R = (pq - qp) + (-pr + rp) + (-q + q) + (r - r) + (qr - rq) + (p - p)$

Sum of the exponents of $R = 0 + 0 + 0 + 0 + 0 + 0 = 0$

So, the term with $R$ is $R^0$.

Therefore, the LHS becomes:

LHS $= A^0 \cdot R^0$

Assuming the G.P. is non-trivial (i.e., $A \neq 0$ and $R \neq 0$), we have $A^0 = 1$ and $R^0 = 1$.

LHS $= 1 \cdot 1 = 1$

Thus, we have shown that LHS = 1.

LHS = RHS

Conclusion:

Hence, it is proved that $a^{q - r} b^{r - p} c^{p - q} = 1$ if the p^th, q^th, and r^th terms of a G.P. are a, b, and c respectively.

Given:

The first term of a G.P. is $a_1 = a$.

The n^th term of the G.P. is $a_n = b$.

The product of the first n terms is $P$.

To Prove:

$P^2 = (ab)^n$.

Proof:

Let the common ratio of the G.P. be $r$.

The terms of the G.P. are $a_1, a_2, a_3, \dots, a_n$.

The terms can be written as $a, ar, ar^2, \dots, ar^{n-1}$.

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

The n^th term is given by the formula $a_n = a_1 r^{n-1}$. Since $a_1 = a$ and $a_n = b$, we have:

$b = ar^{n-1}$

The product of the first n terms is $P = a_1 \times a_2 \times a_3 \times \dots \times a_n$.

Substitute the terms in terms of $a$ and $r$:

$P = (a) \times (ar) \times (ar^2) \times \dots \times (ar^{n-1})$

Group the terms with base $a$ and the terms with base $r$:

$P = (a \cdot a \cdot a \cdot \dots \cdot a \text{ (n times)}) \times (r^0 \cdot r^1 \cdot r^2 \cdot \dots \cdot r^{n-1})$

$P = a^n \cdot r^{(0 + 1 + 2 + \dots + (n-1))}$

The sum of the exponents of $r$ is the sum of the first $(n-1)$ non-negative integers, which is the sum of an arithmetic series with first term 0, last term $(n-1)$, and number of terms $n$. The sum is $\frac{n(0 + (n-1))}{2} = \frac{n(n-1)}{2}$.

$P = a^n r^{\frac{n(n-1)}{2}}$

Now, we need to find $P^2$. Square both sides of the equation for $P$:

$P^2 = \left(a^n r^{\frac{n(n-1)}{2}}\right)^2$

Using the properties of exponents $(xy)^m = x^m y^m$ and $(x^m)^k = x^{mk}$:

$P^2 = (a^n)^2 \left(r^{\frac{n(n-1)}{2}}\right)^2$

$P^2 = a^{2n} r^{2 \times \frac{n(n-1)}{2}}$

$P^2 = a^{2n} r^{n(n-1)}$

Now, consider the right-hand side (RHS) of the equation we need to prove, which is $(ab)^n$.

We are given $a_1 = a$ and $a_n = b$. We also know $b = ar^{n-1}$.

RHS $= (ab)^n$

Substitute the expression for $b$ in terms of $a$ and $r$:

RHS $= (a \cdot (ar^{n-1}))^n$

Simplify the term inside the parenthesis: $a \cdot ar^{n-1} = a^{1+1} r^{n-1} = a^2 r^{n-1}$.

RHS $= (a^2 r^{n-1})^n$

Using the properties of exponents $(xy)^m = x^m y^m$ and $(x^m)^k = x^{mk}$:

RHS $= (a^2)^n (r^{n-1})^n$

RHS $= a^{2n} r^{(n-1)n}$

RHS $= a^{2n} r^{n(n-1)}$

Comparing the expressions for $P^2$ and $(ab)^n$:

$P^2 = a^{2n} r^{n(n-1)}$

$(ab)^n = a^{2n} r^{n(n-1)}$

Since both expressions are equal, $P^2 = (ab)^n$.

Conclusion:

Therefore, it is proved that $P^2 = (ab)^n$, where $a$ is the first term, $b$ is the n^th term, and $P$ is the product of the first n terms of a G.P.

Given:

A Geometric Progression (G.P.).

Sum of the first $n$ terms ($S_n$).

Sum of terms from the $(n+1)^{\text{th}}$ to the $(2n)^{\text{th}}$ term ($S_{n+1 \text{ to } 2n}$).

To Show:

The ratio $\frac{S_n}{S_{n+1 \text{ to } 2n}} = \frac{1}{r^n}$, where $r$ is the common ratio of the G.P.

Solution:

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

The terms of the G.P. are $a, ar, ar^2, \dots, ar^{n-1}, ar^n, ar^{n+1}, \dots, ar^{2n-1}, \dots$

Case 1: $r \neq 1$

The sum of the first $n$ terms of a G.P. is given by the formula:

The terms from the $(n+1)^{\text{th}}$ to the $(2n)^{\text{th}}$ term are $a_{n+1}, a_{n+2}, \dots, a_{2n}$.

These terms are $ar^n, ar^{n+1}, ar^{n+2}, \dots, ar^{2n-1}$.

The sum of these terms is $S_{n+1 \text{ to } 2n} = ar^n + ar^{n+1} + ar^{n+2} + \dots + ar^{2n-1}$.

This is a geometric series with:

First term ($a'$) $= ar^n$

Common ratio ($r'$) $= \frac{ar^{n+1}}{ar^n} = r$

Number of terms ($N$) $= (2n) - (n+1) + 1 = n$.

The sum of this G.P. is given by $S_{N} = \frac{a'((r')^N - 1)}{r' - 1}$.

Substitute the values $a' = ar^n$, $r' = r$, and $N = n$:

Now, consider the ratio of the sum of the first $n$ terms to the sum of terms from the $(n+1)^{\text{th}}$ to the $(2n)^{\text{th}}$ term. Divide equation (i) by equation (ii).

Ratio $= \frac{S_n}{S_{n+1 \text{ to } 2n}} = \frac{\frac{a(r^n - 1)}{r - 1}}{\frac{ar^n(r^n - 1)}{r - 1}}$

Assume $a \neq 0$ (for a non-trivial G.P.) and $r \neq 1$. Also assume $r^n \neq 1$ (otherwise both sums might be zero). With these assumptions, we can cancel the common factors $\frac{a(r^n - 1)}{r-1}$.

Ratio $= \frac{\cancel{\frac{a(r^n - 1)}{r - 1}}}{\cancel{\frac{a(r^n - 1)}{r - 1}} \cdot r^n} = \frac{1}{r^n}$

The ratio is $\frac{1}{r^n}$.

Case 2: $r = 1$

If the common ratio is $r=1$, the G.P. is $a, a, a, \dots$.

The sum of the first $n$ terms is $S_n = a + a + \dots + a$ (n terms) $= na$.

The terms from the $(n+1)^{\text{th}}$ to the $(2n)^{\text{th}}$ term are $a, a, \dots, a$ (n terms).

The sum of these terms is $S_{n+1 \text{ to } 2n} = a + a + \dots + a$ (n terms) $= na$.

The ratio is $\frac{S_n}{S_{n+1 \text{ to } 2n}} = \frac{na}{na} = 1$.

The formula $\frac{1}{r^n}$ for $r=1$ gives $\frac{1}{1^n} = \frac{1}{1} = 1$.

The formula holds for $r=1$ as well.

In both cases ($r \neq 1$ and $r=1$), the ratio of the sum of the first $n$ terms to the sum of terms from the $(n+1)^{\text{th}}$ to the $(2n)^{\text{th}}$ term is $\frac{1}{r^n}$, assuming $a \neq 0$.

Conclusion:

Thus, it is shown that the ratio of the sum of the first $n$ terms of a G.P. to the sum of terms from the $(n+1)^{\text{th}}$ to the $(2n)^{\text{th}}$ term is $\frac{1}{r^n}$.

Given:

a, b, c, and d are in Geometric Progression (G.P.).

To Show:

$(a^2 + b^2 + c^2 ) (b^2 + c^2 + d^2 ) = (ab + bc + cd)^2$.

Solution:

Since a, b, c, and d are in G.P., there exists a common ratio, let's denote it by $r$.

Then, we can express b, c, and d in terms of a and r:

$b = ar$

$c = ar^2$

$d = ar^3$

Consider the left-hand side (LHS) of the equation:

LHS $= (a^2 + b^2 + c^2 ) (b^2 + c^2 + d^2 )$

Substitute the expressions for b, c, and d:

LHS $= (a^2 + (ar)^2 + (ar^2)^2 ) ( (ar)^2 + (ar^2)^2 + (ar^3)^2 )$

LHS $= (a^2 + a^2r^2 + a^2r^4 ) ( a^2r^2 + a^2r^4 + a^2r^6 )$

Factor out common terms from each parenthesis:

LHS $= a^2(1 + r^2 + r^4 ) \cdot a^2r^2(1 + r^2 + r^4 )$

Multiply the terms:

LHS $= a^2 \cdot a^2r^2 \cdot (1 + r^2 + r^4) \cdot (1 + r^2 + r^4)$

LHS $= a^{2+2}r^2 (1 + r^2 + r^4)^2$

LHS $= a^4r^2 (1 + r^2 + r^4)^2$

Now, consider the right-hand side (RHS) of the equation:

RHS $= (ab + bc + cd)^2$

Substitute the expressions for b, c, and d:

RHS $= (a(ar) + (ar)(ar^2) + (ar^2)(ar^3))^2$

Multiply the terms inside the parenthesis:

$ab = a \cdot ar = a^2r$

$bc = ar \cdot ar^2 = a^2r^{1+2} = a^2r^3$

$cd = ar^2 \cdot ar^3 = a^2r^{2+3} = a^2r^5$

So, the expression inside the parenthesis is:

$ab + bc + cd = a^2r + a^2r^3 + a^2r^5$

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

$ab + bc + cd = a^2r(1 + r^2 + r^4)$

Now, square this expression to get the RHS:

RHS $= [a^2r(1 + r^2 + r^4)]^2$

Using the property $(xyz)^m = x^m y^m z^m$:

RHS $= (a^2r)^2 (1 + r^2 + r^4)^2$

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

RHS $= (a^2)^2 r^2 (1 + r^2 + r^4)^2$

RHS $= a^{2 \times 2} r^2 (1 + r^2 + r^4)^2$

RHS $= a^4r^2 (1 + r^2 + r^4)^2$

Comparing the simplified expressions for LHS and RHS:

LHS $= a^4r^2 (1 + r^2 + r^4)^2$

RHS $= a^4r^2 (1 + r^2 + r^4)^2$

Since LHS = RHS, the identity is proven.

This proof holds even if $a=0$ (all terms are 0, both sides are 0) or $r=0$ (terms $a, 0, 0, 0$, both sides are 0) or if any of the terms are zero, provided they form a G.P.

Conclusion:

Thus, it is shown that if a, b, c, and d are in G.P., then $(a^2 + b^2 + c^2 ) (b^2 + c^2 + d^2 ) = (ab + bc + cd)^2$.

Given:

The first number is 3.

The last number is 81.

Two numbers are to be inserted between 3 and 81 to form a G.P.

To Find:

The two numbers that, when inserted between 3 and 81, form a G.P.

Solution:

Let the two numbers to be inserted between 3 and 81 be $G_1$ and $G_2$.

The resulting sequence in G.P. will be $3, G_1, G_2, 81$.

This sequence has 4 terms.

The first term is $a_1 = 3$.

The fourth term is $a_4 = 81$.

The formula for the n^th term of a G.P. is $a_n = a_1 r^{n-1}$, where $r$ is the common ratio.

For the 4th term ($n=4$):

$a_4 = a_1 r^{4-1}$

$a_4 = a_1 r^3$

Substitute the given values $a_1 = 3$ and $a_4 = 81$:

$81 = 3 \cdot r^3$

Divide both sides by 3 to solve for $r^3$:

$\frac{81}{3} = r^3$

$27 = r^3$

Take the cube root of both sides to find the common ratio $r$:

$r = \sqrt[3]{27}$

$r = 3$

Since we are looking for real numbers in the G.P., we consider the real cube root.

Now we find the two inserted numbers using the first term $a_1 = 3$ and the common ratio $r = 3$.

The inserted numbers are the second term ($a_2$) and the third term ($a_3$) of the G.P.

The second term ($G_1$):

$G_1 = a_2 = a_1 r^{2-1} = a_1 r$

$G_1 = 3 \times 3 = 9$

The third term ($G_2$):

$G_2 = a_3 = a_1 r^{3-1} = a_1 r^2$

$G_2 = 3 \times 3^2 = 3 \times 9 = 27$

The resulting sequence is 3, 9, 27, 81.

Verification:

Check the ratio between consecutive terms:

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

$\frac{27}{9} = 3$

$\frac{81}{27} = 3$

The common ratio is indeed 3, so the sequence is a G.P.

Final Answer:

The two numbers to be inserted between 3 and 81 are 9 and 27.

Given:

The expression $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$.

The geometric mean between $a$ and $b$.

To Find:

The value of $n$ such that the given expression is equal to the geometric mean between $a$ and $b$.

Solution:

The geometric mean (G.M.) between two positive numbers $a$ and $b$ is $\sqrt{ab}$.

We are given that the expression $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is equal to the geometric mean between $a$ and $b$. So, we set up the equation:

We can write $\sqrt{ab}$ using exponents as $(ab)^{\frac{1}{2}} = a^{\frac{1}{2}} b^{\frac{1}{2}}$. Substitute this into equation (i):

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

Cross-multiply to remove the fraction:

$a^{n+1}+b^{n+1} = a^{\frac{1}{2}}b^{\frac{1}{2}}(a^n+b^n)$

Distribute the term $a^{\frac{1}{2}}b^{\frac{1}{2}}$ on the right side:

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

Using the exponent rule $x^m \cdot x^p = x^{m+p}$, combine the terms with the same base:

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

Rearrange the terms by grouping terms with base $a$ on one side and terms with base $b$ on the other side. Move terms from the right side to the left side:

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

Factor out the common terms from each side. On the left side, the common term is $a^{n + \frac{1}{2}}$ (since $n+1 = n + \frac{1}{2} + \frac{1}{2}$).

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

Simplify the exponent $(n+1) - (n + \frac{1}{2}) = \frac{1}{2}$:

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

On the right side, the common term is $b^{n + \frac{1}{2}}$:

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

Move all terms to the left side:

$a^{n + \frac{1}{2}}(a^{\frac{1}{2}} - b^{\frac{1}{2}}) - b^{n + \frac{1}{2}}(a^{\frac{1}{2}} - b^{\frac{1}{2}}) = 0$

Factor out the common binomial term $(a^{\frac{1}{2}} - b^{\frac{1}{2}})$:

$(a^{\frac{1}{2}} - b^{\frac{1}{2}})(a^{n + \frac{1}{2}} - b^{n + \frac{1}{2}}) = 0$

For the product of two factors to be zero, at least one of the factors must be zero.

Case 1: $a^{\frac{1}{2}} - b^{\frac{1}{2}} = 0$

$a^{\frac{1}{2}} = b^{\frac{1}{2}}$

Squaring both sides gives $a = b$. If $a=b$, the geometric mean is $\sqrt{a \cdot a} = a$, and the original expression is $\frac{a^{n+1}+a^{n+1}}{a^n+a^n} = \frac{2a^{n+1}}{2a^n} = a$. So the equality holds for any $n$ if $a=b$. However, the question implies finding a specific value of $n$ for general $a$ and $b$, suggesting $a \neq b$. If $a \neq b$, then $a^{\frac{1}{2}} - b^{\frac{1}{2}} \neq 0$.

Case 2: $a^{n + \frac{1}{2}} - b^{n + \frac{1}{2}} = 0$

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

Assuming $b \neq 0$, divide both sides by $b^{n + \frac{1}{2}}$:

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

Using the exponent rule $\frac{x^m}{y^m} = \left(\frac{x}{y}\right)^m$:

$\left(\frac{a}{b}\right)^{n + \frac{1}{2}} = 1$

For a positive base $\left(\frac{a}{b}\right)$ not equal to 1 (since we assumed $a \neq b$), the only way this expression can be equal to 1 is if the exponent is 0.

Subtract $\frac{1}{2}$ from both sides:

Final Answer:

The value of n is $-\frac{1}{2}$.

Given:

Let the two positive numbers be $a$ and $b$.

The sum of the numbers is $a+b$.

The geometric mean (G.M.) of the numbers is $\sqrt{ab}$.

The given condition is $a+b = 6\sqrt{ab}$.

To Show:

The numbers $a$ and $b$ are in the ratio $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$.

Solution:

We are given the relationship between the sum and the geometric mean of the two numbers $a$ and $b$:

$a + b = 6\sqrt{ab}$

To eliminate the square root, square both sides of the equation:

$(a + b)^2 = (6\sqrt{ab})^2$

Expand the left side and square the right side:

$a^2 + 2ab + b^2 = 36ab$

Subtract $2ab$ from both sides of the equation:

$a^2 + b^2 = 36ab - 2ab$

$a^2 + b^2 = 34ab$

To find the ratio of $a$ to $b$, divide the entire equation by $b^2$ (assuming $b \neq 0$; if $b=0$, then $a+0 = 6\sqrt{a \cdot 0} \implies a=0$, which would mean the numbers are 0 and 0, and the ratio is undefined. The problem implies positive numbers for a positive geometric mean).

$\frac{a^2}{b^2} + \frac{b^2}{b^2} = \frac{34ab}{b^2}$

$\left(\frac{a}{b}\right)^2 + 1 = 34 \left(\frac{a}{b}\right)$

Let $x = \frac{a}{b}$. The equation becomes a quadratic equation in terms of $x$:

$x^2 + 1 = 34x$

Rearrange the terms to the standard quadratic form $Ax^2 + Bx + C = 0$:

$x^2 - 34x + 1 = 0$

We can solve this quadratic equation for $x$ using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$. Here, $A=1$, $B=-34$, and $C=1$.

$x = \frac{-(-34) \pm \sqrt{(-34)^2 - 4(1)(1)}}{2(1)}$

$x = \frac{34 \pm \sqrt{1156 - 4}}{2}$

$x = \frac{34 \pm \sqrt{1152}}{2}$

Simplify the square root $\sqrt{1152}$. We can find the prime factorization of 1152 or recognize that $1152 = 2 \times 576 = 2 \times 24^2$.

$\sqrt{1152} = \sqrt{2 \times 24^2} = 24\sqrt{2}$

Substitute this back into the expression for $x$:

$x = \frac{34 \pm 24\sqrt{2}}{2}$

Divide both terms in the numerator by 2:

$x = \frac{34}{2} \pm \frac{24\sqrt{2}}{2}$

$x = 17 \pm 12\sqrt{2}$

So, the ratio $\frac{a}{b}$ can be either $17 + 12\sqrt{2}$ or $17 - 12\sqrt{2}$.

We need to show that this ratio is equivalent to $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$. This means we need to show that either $\frac{a}{b} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$ or $\frac{a}{b} = \frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}}$.

Let's evaluate the ratio $\frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$ by rationalizing the denominator. Multiply the numerator and denominator by the conjugate of the denominator, which is $(3 + 2\sqrt{2})$.

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

Using the formulas $(x+y)^2 = x^2 + 2xy + y^2$ and $(x-y)(x+y) = x^2 - y^2$:

Numerator $= (3 + 2\sqrt{2})^2 = 3^2 + 2(3)(2\sqrt{2}) + (2\sqrt{2})^2 = 9 + 12\sqrt{2} + (4 \times 2) = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2}$

Denominator $= (3 - 2\sqrt{2})(3 + 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - (4 \times 2) = 9 - 8 = 1$

So, $\frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{17 + 12\sqrt{2}}{1} = 17 + 12\sqrt{2}$.

This matches one of the possible values for $\frac{a}{b}$.

The other possible value for $\frac{a}{b}$ is $17 - 12\sqrt{2}$. Let's check the reciprocal ratio $\frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}}$:

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

Numerator $= (3 - 2\sqrt{2})^2 = 3^2 - 2(3)(2\sqrt{2}) + (2\sqrt{2})^2 = 9 - 12\sqrt{2} + 8 = 17 - 12\sqrt{2}$

Denominator $= 1$

So, $\frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} = 17 - 12\sqrt{2}$.

The possible values for the ratio $\frac{a}{b}$ are exactly the values $\frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$ and $\frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}}$.

This means that the ratio $a:b$ is either $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$ or $(3 - 2\sqrt{2}) : (3 + 2\sqrt{2})$. In either case, the numbers are in the ratio $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$ (since the order of the numbers $a$ and $b$ can be swapped).

Conclusion:

Hence, it is shown that if the sum of two numbers is 6 times their geometric mean, the numbers are in the ratio $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$.

Given:

Let the two positive numbers be $x$ and $y$.

Arithmetic Mean (A.M.) between $x$ and $y$ is $A$.

Geometric Mean (G.M.) between $x$ and $y$ is $G$.

To Prove:

The two numbers are $A + \sqrt{(A + G) (A - G)}$ and $A - \sqrt{(A + G) (A - G)}$.

Proof:

By the definitions of A.M. and G.M., we have:

From equation (i), we get the sum of the numbers:

$x+y = 2A$

From equation (ii), square both sides to get the product of the numbers:

$(\sqrt{xy})^2 = G^2$

$xy = G^2$

We have the sum ($x+y$) and the product ($xy$) of two numbers. These two numbers are the roots of a quadratic equation $t^2 - (\text{sum})t + (\text{product}) = 0$.

The quadratic equation whose roots are $x$ and $y$ is:

$t^2 - (x+y)t + xy = 0$

Substitute the values from (iii) and (iv):

$t^2 - (2A)t + G^2 = 0$

We can solve this quadratic equation for $t$ using the quadratic formula $t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$. Here, the coefficients are $A_{\text{quad}}=1$, $B_{\text{quad}}=-2A$, and $C_{\text{quad}}=G^2$.

$t = \frac{-(-2A) \pm \sqrt{(-2A)^2 - 4(1)(G^2)}}{2(1)}$

$t = \frac{2A \pm \sqrt{4A^2 - 4G^2}}{2}$

Factor out 4 from the term under the square root:

$t = \frac{2A \pm \sqrt{4(A^2 - G^2)}}{2}$

Take the square root of 4:

$t = \frac{2A \pm 2\sqrt{A^2 - G^2}}{2}$

Divide both terms in the numerator by 2:

$t = \frac{2A}{2} \pm \frac{2\sqrt{A^2 - G^2}}{2}$

$t = A \pm \sqrt{A^2 - G^2}$

Using the difference of squares formula $A^2 - G^2 = (A - G)(A + G)$, we can rewrite the term under the square root:

$t = A \pm \sqrt{(A - G)(A + G)}$

The two roots of this equation are the two numbers $x$ and $y$.

So, the two numbers are $A + \sqrt{(A - G)(A + G)}$ and $A - \sqrt{(A - G)(A + G)}$.

Since the product $(A+G)(A-G)$ is the same as $(A-G)(A+G)$, the expression can be written as $A \pm \sqrt{(A + G) (A - G)}$.

For the numbers to be real, we must have $A^2 - G^2 \geq 0$, which means $A^2 \geq G^2$. Since A and G are AM and GM of positive numbers, $A \ge G$, which implies $A^2 \ge G^2$. For the numbers to be positive, $A > \sqrt{A^2 - G^2}$, which is true if $G > 0$. If $G=0$, then the numbers must be 0, but the problem states positive numbers. So $G>0$ is implied.

Conclusion:

Therefore, it is proved that the two positive numbers are $A \pm \sqrt{(A + G) (A - G)}$.

Given:

Initial number of bacteria at the start (hour 0) $= 30$.

The number of bacteria doubles every hour.

To Find:

The number of bacteria present at the end of the 2^nd hour, 4^th hour, and n^th hour.

Solution:

Let the number of bacteria at the end of $t$ hours be denoted by $N_t$.

The number of bacteria originally present (at the start, i.e., $t=0$) is $N_0 = 30$.

Since the number of bacteria doubles every hour, the common ratio of growth is $r = 2$.

The number of bacteria at the end of each hour forms a geometric progression starting from $N_0$.

At the end of 1^st hour ($t=1$): $N_1 = N_0 \times r = 30 \times 2^1 = 60$.

At the end of 2^nd hour ($t=2$): $N_2 = N_1 \times r = (30 \times 2) \times 2 = 30 \times 2^2$.

At the end of 3^rd hour ($t=3$): $N_3 = N_2 \times r = (30 \times 2^2) \times 2 = 30 \times 2^3$.

In general, the number of bacteria at the end of the $t^{\text{th}}$ hour is given by $N_t = N_0 \times r^t$.

Substitute the initial number $N_0 = 30$ and the growth ratio $r = 2$:

$N_t = 30 \times 2^t$

Now we calculate the number of bacteria for the required times:

Number of bacteria at the end of 2^nd hour ($t=2$):

$N_2 = 30 \times 2^2$

$N_2 = 30 \times 4$

$N_2 = 120$

Number of bacteria at the end of 4^th hour ($t=4$):

$N_4 = 30 \times 2^4$

$N_4 = 30 \times 16$

$N_4 = 480$

Number of bacteria at the end of n^th hour ($t=n$):

$N_n = 30 \times 2^n$

Final Answer:

The number of bacteria present at the end of the 2^nd hour is 120.

The number of bacteria present at the end of the 4^th hour is 480.

The number of bacteria present at the end of the n^th hour is $30 \times 2^n$.

Given:

Principal amount (P) $= \textsf{₹}500$

Time period (t) $= 10$ years

Annual interest rate (r) $= 10\%$ per annum $= 0.10$

Compounding frequency = annually (n = 1)

To Find:

The amount (A) after 10 years when compounded annually.

Solution:

The formula for the amount A with compound interest is given by:

$A = P(1 + \frac{r}{n})^{nt}$

Since the interest is compounded annually, the number of times interest is compounded per year is $n = 1$. The formula simplifies to:

$A = P(1 + r)^t$

Substitute the given values into the formula:

$P = 500$, $r = 0.10$, and $t = 10$.

$A = 500(1 + 0.10)^{10}$

$A = 500(1.1)^{10}$

We need to calculate $(1.1)^{10}$.

$(1.1)^{10} \approx 2.5937424601$

Now, substitute this value back into the equation for A:

$A \approx 500 \times 2.5937424601$

Perform the multiplication:

$A \approx 1296.87123005$

Rounding the amount to two decimal places (as it represents currency):

$A \approx 1296.87$

The amount will be approximately $\textsf{₹}1296.87$ after 10 years.

Final Answer:

The amount after 10 years will be approximately $\textsf{₹}1296.87$.

Given:

Let the roots of the quadratic equation be $\alpha$ and $\beta$.

Arithmetic Mean (A.M.) of the roots = 8

Geometric Mean (G.M.) of the roots = 5

To Obtain:

The quadratic equation.

Solution:

The Arithmetic Mean (A.M.) of two numbers $\alpha$ and $\beta$ is given by $\frac{\alpha + \beta}{2}$.

The Geometric Mean (G.M.) of two non-negative numbers $\alpha$ and $\beta$ is given by $\sqrt{\alpha \beta}$.

Given that the A.M. of the roots is 8:

From equation (1), we find the sum of the roots:

Given that the G.M. of the roots is 5:

Squaring both sides of equation (3), we find the product of the roots:

A quadratic equation with roots $\alpha$ and $\beta$ can be written in the form:

Substituting the values from equation (2) and equation (4) into equation (5):

Thus, the required quadratic equation is:

Final Answer:

The quadratic equation is $\mathbf{x^2 - 16x + 25 = 0}$.

Given:

a, b, c, d and p are different real numbers.

The inequality $(a^2 + b^2 + c^2)p^2 – 2(ab + bc + cd) p + (b^2 + c^2 + d^2 ) \leq 0$ holds.

To Show:

a, b, c and d are in G.P.

(i.e., to show that $\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$ for some common ratio).

Solution:

The given inequality is:

Expand the terms in the inequality:

Rearrange the terms to group them as perfect squares:

Using the algebraic identity $(x-y)^2 = x^2 - 2xy + y^2$, we can rewrite the expression as a sum of squares:

We know that the square of any real number is non-negative. Therefore,

$(ap - b)^2 \geq 0$

$(bp - c)^2 \geq 0$

$(cp - d)^2 \geq 0$

The sum of three non-negative quantities can be less than or equal to zero only if each quantity is equal to zero.

Therefore, we must have:

From equation (2):

From equation (3):

From equation (4):

Since a, b, c, d are different real numbers, $a \neq 0$, $b \neq 0$, $c \neq 0$, $d \neq 0$, and $p \neq 0$ (otherwise, if $p=0$, then $b=c=d=0$, contradicting that they are different). Also $p \neq 1$ (otherwise $a=b=c=d$).

From (5), we have $\frac{b}{a} = p$.

From (6), substituting b from (5), we get c = (ap)p = ap$^2$. Then $\frac{c}{b} = \frac{ap^2}{ap} = p$.

From (7), substituting c from $c=ap^2$, we get d = (ap$^2$)p = ap$^3$. Then $\frac{d}{c} = \frac{ap^3}{ap^2} = p$.

Since $\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = p$, where p is a common ratio, the numbers a, b, c, and d are in Geometric Progression (G.P.).

Final Answer:

Since we have shown that $\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = p$ from the given inequality, a, b, c, and d are in G.P.

Given:

A function f satisfying $f(x + y) = f(x) f(y)$ for all $x, y \in N$.

$f(1) = 3$.

$\sum\limits_{x=1}^{n} f(x) = 120$.

To Find:

The value of n.

Solution:

The given property $f(x + y) = f(x) f(y)$ is the characteristic property of an exponential function of the form $f(x) = a^x$ for some constant a.

Let's verify this using the given information:

$f(1) = a^1 = a$. Since $f(1) = 3$, we have $a = 3$.

So, the function is $f(x) = 3^x$.

Let's check the property: $f(x+y) = 3^{x+y}$ and $f(x)f(y) = 3^x \cdot 3^y = 3^{x+y}$. The property holds.

The given summation is $\sum\limits_{x=1}^{n} f(x) = \sum\limits_{x=1}^{n} 3^x$.

This is the sum of the first n terms of a geometric series with the first term $a_1 = f(1) = 3^1 = 3$ and the common ratio $r = \frac{f(2)}{f(1)} = \frac{3^2}{3^1} = 3$.

The sum of the first n terms of a geometric series is given by the formula $S_n = \frac{a_1(r^n - 1)}{r - 1}$.

Using the given values, $S_n = \frac{3(3^n - 1)}{3 - 1} = \frac{3(3^n - 1)}{2}$.

We are given that the sum is 120.

Multiply both sides of equation (1) by 2:

Divide both sides by 3:

Add 1 to both sides:

We need to express 81 as a power of 3. $81 = 3 \times 3 \times 3 \times 3 = 3^4$.

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

Final Answer:

The value of n is 4.

Given:

Sum of some terms of a G.P., $S_n = 315$.

First term, $a = 5$.

Common ratio, $r = 2$.

To Find:

The last term ($a_n$).

The number of terms (n).

Solution:

The formula for the sum of the first n terms of a geometric progression (G.P.) is given by:

Substitute the given values $S_n = 315$, $a = 5$, and $r = 2$ into equation (1):

Divide both sides by 5:

Add 1 to both sides:

We need to express 64 as a power of 2. $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$.

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

The number of terms is 6.

Now, we need to find the last term, which is the 6th term ($a_6$).

The formula for the n-th term of a G.P. is given by:

Substitute $a = 5$, $r = 2$, and $n = 6$ into equation (2):

Calculate $2^5$:

Substitute the value of $2^5$ back into the equation for $a_6$:

Multiply 5 by 32:

The last term is 160.

Final Answer:

The number of terms is 6.

The last term is 160.

Given:

The first term of the G.P. is $a = 1$.

The sum of the third term ($a_3$) and the fifth term ($a_5$) is 90.

To Find:

The common ratio (r) of the G.P.

Solution:

The formula for the n-th term of a geometric progression is given by:

Using this formula, the third term ($a_3$) is:

Since $a = 1$, we have:

The fifth term ($a_5$) is:

Since $a = 1$, we have:

Now, substitute the expressions for $a_3$ and $a_5$ into equation (1):

Rearrange the equation into a standard form:

This is a quadratic equation in terms of $r^2$. Let $y = r^2$. The equation becomes:

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.

This gives two possible values for y:

$y + 10 = 0 \implies y = -10$

$y - 9 = 0 \implies y = 9$

Now substitute back $y = r^2$:

Case 1: $r^2 = -10$. Since the common ratio r of a G.P. is typically a real number, $r^2$ cannot be negative for a real r. Therefore, this solution is rejected for a G.P. with real terms.

Case 2: $r^2 = 9$.

Taking the square root of both sides:

Thus, the possible values for the common ratio are 3 and -3.

Final Answer:

The common ratio of the G.P. can be 3 or -3.

Given:

Three numbers are in G.P.

The sum of these three numbers is 56.

If 1, 7, and 21 are subtracted from these numbers in that order, the resulting numbers form an A.P.

To Find:

The three numbers in the original G.P.

Solution:

Let the three numbers in G.P. be represented as $\frac{a}{r}$, $a$, and $ar$, where $a$ is the middle term and $r$ is the common ratio.

According to the first condition, the sum of these numbers is 56:

The numbers obtained after subtracting 1, 7, and 21 from these terms are $(\frac{a}{r} - 1)$, $(a - 7)$, and $(ar - 21)$.

According to the second condition, these three numbers form an A.P.

In an A.P., the middle term is the average of the first and third terms, or equivalently, the difference between consecutive terms is constant.

Using the property that the difference between consecutive terms is equal:

Simplify equation (2):

Rearrange the terms:

Multiply by r (assuming $r \neq 0$, which must be true for a G.P. with non-zero sum):

Factor out 'a':

From equation (1), factor out 'a':

From equation (3), assuming $r \neq 1$, we can write $a = \frac{8r}{(r-1)^2}$. Substitute this into equation (4):

Since $a, b, c, d$ are real numbers and the sum is positive, and $a = \frac{8r}{(r-1)^2}$, $r$ must be positive. Assuming $r \neq 0$, we can divide both sides by $8r$:

Multiply both sides by $(r-1)^2$:

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

Divide the equation by 3:

Factor the quadratic equation:

This gives two possible values for the common ratio r:

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

$r - 2 = 0 \implies r = 2$

Now, find the value of 'a' for each value of r using equation (3): $a(r-1)^2 = 8r$.

Case 1: $r = 2$

The G.P. terms are $\frac{a}{r}, a, ar$:

The numbers are 8, 16, 32. Their sum is $8 + 16 + 32 = 56$.

Subtracting 1, 7, 21: $8-1, 16-7, 32-21 = 7, 9, 11$. This is an A.P. with common difference 2.

Case 2: $r = \frac{1}{2}$

The G.P. terms are $\frac{a}{r}, a, ar$:

The numbers are 32, 16, 8. Their sum is $32 + 16 + 8 = 56$.

Subtracting 1, 7, 21: $32-1, 16-7, 8-21 = 31, 9, -13$. This is an A.P. with common difference $9 - 31 = -22$ and $-13 - 9 = -22$.

Both sets of numbers satisfy the given conditions.

Final Answer:

The three numbers are either 8, 16, 32 or 32, 16, 8.

Given:

A G.P. with an even number of terms.

Let the number of terms be $2n$, where $n$ is a positive integer.

Let the first term be $a$ and the common ratio be $r$. The terms of the G.P. are $a, ar, ar^2, \dots, ar^{2n-1}$.

The sum of all the terms ($S_{2n}$) is 5 times the sum of the terms occupying odd places ($S_{odd}$).

To Find:

The common ratio (r) of the G.P.

Solution:

The sum of all $2n$ terms of the G.P. is given by the formula:

Note that this formula is valid for $r \neq 1$. If $r=1$, the G.P. is $a, a, \dots, a$, and $S_{2n} = 2na$. The terms in odd places are $a, a, \dots, a$ (n terms), $S_{odd} = na$. The condition $S_{2n} = 5 S_{odd}$ becomes $2na = 5na$, which implies $2=5$ (since $a \neq 0$ and $n \ge 1$), which is false. So $r \neq 1$.

The terms occupying odd places are the 1st, 3rd, 5th, ..., $(2n-1)$-th terms.

These terms are $a \cdot r^{1-1}, a \cdot r^{3-1}, a \cdot r^{5-1}, \dots, a \cdot r^{(2n-1)-1}$.

So the terms are $a, ar^2, ar^4, \dots, ar^{2n-2}$.

This sequence of terms also forms a G.P. with:

First term = $A = a$

Common ratio = $R = \frac{ar^2}{a} = r^2$

Number of terms = There are $n$ terms in the sum (from index 1 to $2n-1$, taking every other term: $1, 3, \dots, 2n-1$, which are $n$ terms).

The sum of these terms ($S_{odd}$) is given by the formula for a G.P. sum with these parameters:

This formula is valid for $r^2 \neq 1$, i.e., $r \neq \pm 1$. We've already shown $r \neq 1$. If $r=-1$, $S_{2n}=0$ (sum of $2n$ terms of $a, -a, a, -a, \dots$). $S_{odd} = na$ (sum of $n$ terms of $a, a, \dots$). The condition $0 = 5na$ implies $a=0$ (since $n \ge 1$), which means all terms are zero, a trivial case. So we assume $a \neq 0$ and thus $r \neq -1$.

Now, substitute the expressions for $S_{2n}$ from (2) and $S_{odd}$ from (3) into the given condition (1):

Since $a \neq 0$ and $r \neq \pm 1$ (which implies $r^{2n} - 1 \neq 0$ for $n \ge 1$), we can cancel $a$ and $(r^{2n} - 1)$ from both sides:

Use the identity $r^2 - 1 = (r - 1)(r + 1)$:

Multiply both sides by $(r - 1)(r + 1)$. Since $r \neq 1$, $r - 1 \neq 0$. Since $r \neq -1$, $r + 1 \neq 0$.

Solve for r:

Final Answer:

The common ratio of the G.P. is 4.

Given:

The equality $\frac{a \;+\; bx}{a \;-\; bx}$ = $\frac{b \;+\; cx}{b \;-\; cx}$ = $\frac{c \;+\; dx}{c \;-\; dx}$, where $x \neq 0$.

To Prove:

a, b, c, and d are in Geometric Progression (G.P.).

Solution:

Let the given equalities be:

$\frac{a \;+\; bx}{a \;-\; bx}$ = $\frac{b \;+\; cx}{b \;-\; cx}$

and

$\frac{b \;+\; cx}{b \;-\; cx}$ = $\frac{c \;+\; dx}{c \;-\; dx}$

We will apply the property of Componendo and Dividendo to the first equality. The property states that if $\frac{m}{n} = \frac{p}{q}$, then $\frac{m+n}{m-n} = \frac{p+q}{p-q}$.

Applying Componendo and Dividendo to the first equality $\frac{a \;+\; bx}{a \;-\; bx}$ = $\frac{b \;+\; cx}{b \;-\; cx}$:

$\frac{(a \;+\; bx) + (a \;-\; bx)}{(a \;+\; bx) - (a \;-\; bx)} = \frac{(b \;+\; cx) + (b \;-\; cx)}{(b \;+\; cx) - (b \;-\; cx)}$

Simplifying the terms:

$\frac{a + bx + a - bx}{a + bx - a + bx} = \frac{b + cx + b - cx}{b + cx - b + cx}$

$\frac{2a}{2bx} = \frac{2b}{2cx}$

$\frac{a}{bx} = \frac{b}{cx}$

Since $x \neq 0$, we can cancel $x$ from both denominators:

$\frac{a}{b} = \frac{b}{c}$

Cross-multiplying, we get:

Now, applying Componendo and Dividendo to the second equality $\frac{b \;+\; cx}{b \;-\; cx}$ = $\frac{c \;+\; dx}{c \;-\; dx}$:

$\frac{(b \;+\; cx) + (b \;-\; cx)}{(b \;+\; cx) - (b \;-\; cx)} = \frac{(c \;+\; dx) + (c \;-\; dx)}{(c \;+\; dx) - (c \;-\; dx)}$

Simplifying the terms:

$\frac{b + cx + b - cx}{b + cx - b + cx} = \frac{c + dx + c - dx}{c + dx - c + dx}$

$\frac{2b}{2cx} = \frac{2c}{2dx}$

$\frac{b}{cx} = \frac{c}{dx}$

Since $x \neq 0$, we can cancel $x$ from both denominators:

$\frac{b}{c} = \frac{c}{d}$

Cross-multiplying, we get:

Conclusion:

From equation (i), $b^2 = ac$, which can be written as $\frac{b}{a} = \frac{c}{b}$ (assuming $a, b, c \neq 0$). This shows that a, b, and c are in G.P.

From equation (ii), $c^2 = bd$, which can be written as $\frac{c}{b} = \frac{d}{c}$ (assuming $b, c, d \neq 0$). This shows that b, c, and d are in G.P.

Since $\frac{b}{a} = \frac{c}{b}$ and $\frac{c}{b} = \frac{d}{c}$, it follows that $\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$.

Therefore, a, b, c, and d are in Geometric Progression.

Given:

S is the sum of n terms in a Geometric Progression (G.P.).

P is the product of n terms in a G.P.

R is the sum of the reciprocals of n terms in a G.P.

To Prove:

$P^2R^n = S^n$

Solution:

Let the Geometric Progression be $a, ar, ar^2, \dots, ar^{n-1}$, where $a$ is the first term and $r$ is the common ratio.

The sum of the n terms (S) is given by:

$S = a + ar + ar^2 + \dots + ar^{n-1}$

If $r \neq 1$, the formula for the sum is $S = a \frac{r^n - 1}{r - 1}$.

If $r = 1$, the G.P. is $a, a, \dots, a$ (n terms), so $S = na$.

The product of the n terms (P) is given by:

$P = a \times ar \times ar^2 \times \dots \times ar^{n-1}$

$P = a^n \times r^{0+1+2+\dots+(n-1)}$

The sum of the exponents of r is the sum of an arithmetic series from 0 to $n-1$, which is $\frac{(n-1)(0 + n-1)}{2} = \frac{n(n-1)}{2}$.

So, $P = a^n r^{\frac{n(n-1)}{2}}$.

The reciprocals of the n terms are $\frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \dots, \frac{1}{ar^{n-1}}$.

This sequence of reciprocals also forms a G.P. with the first term $a' = \frac{1}{a}$ and the common ratio $r' = \frac{1}{r}$.

The sum of the reciprocals (R) is the sum of this new G.P.

If $r \neq 1$ (which means $r' = \frac{1}{r} \neq 1$), the formula for the sum R is:

$R = a' \frac{(r')^n - 1}{r' - 1} = \frac{1}{a} \frac{(\frac{1}{r})^n - 1}{\frac{1}{r} - 1}$

$R = \frac{1}{a} \frac{\frac{1^n - r^n}{r^n}}{\frac{1 - r}{r}} = \frac{1}{a} \frac{1 - r^n}{r^n} \times \frac{r}{1 - r}$

$R = \frac{1}{a} \frac{-(r^n - 1)}{r^{n-1}(r)(-(r - 1)/r)} = \frac{1}{a} \frac{-(r^n - 1)}{r^{n-1}(-(r - 1))}$

$R = \frac{1}{ar^{n-1}} \frac{r^n - 1}{r - 1}$.

If $r = 1$, the reciprocals are $\frac{1}{a}, \frac{1}{a}, \dots, \frac{1}{a}$ (n terms). So, $R = n \times \frac{1}{a} = \frac{n}{a}$.

Now we need to prove $P^2R^n = S^n$. Let's consider the case where $r \neq 1$ first.

Left Hand Side (LHS) = $P^2R^n$

$P^2 = \left(a^n r^{\frac{n(n-1)}{2}}\right)^2 = (a^n)^2 (r^{\frac{n(n-1)}{2}})^2 = a^{2n} r^{n(n-1)}$.

$R^n = \left(\frac{1}{ar^{n-1}} \frac{r^n - 1}{r - 1}\right)^n = \left(\frac{1}{ar^{n-1}}\right)^n \left(\frac{r^n - 1}{r - 1}\right)^n = \frac{1}{a^n (r^{n-1})^n} \left(\frac{r^n - 1}{r - 1}\right)^n = \frac{1}{a^n r^{n(n-1)}} \left(\frac{r^n - 1}{r - 1}\right)^n$.

Now substitute $P^2$ and $R^n$ into the LHS:

LHS = $\left(a^{2n} r^{n(n-1)}\right) \times \left(\frac{1}{a^n r^{n(n-1)}} \left(\frac{r^n - 1}{r - 1}\right)^n\right)$

LHS = $a^{2n} \times r^{n(n-1)} \times \frac{1}{a^n} \times \frac{1}{r^{n(n-1)}} \times \left(\frac{r^n - 1}{r - 1}\right)^n$

LHS = $\frac{a^{2n}}{a^n} \times \frac{r^{n(n-1)}}{r^{n(n-1)}} \times \left(\frac{r^n - 1}{r - 1}\right)^n$

LHS = $a^{2n-n} \times r^{n(n-1)-n(n-1)} \times \left(\frac{r^n - 1}{r - 1}\right)^n$

LHS = $a^n \times r^0 \times \left(\frac{r^n - 1}{r - 1}\right)^n$

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

Right Hand Side (RHS) = $S^n$

Substitute the formula for S (when $r \neq 1$):

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

Comparing the LHS and RHS for $r \neq 1$:

LHS = $a^n \left(\frac{r^n - 1}{r - 1}\right)^n$

RHS = $a^n \left(\frac{r^n - 1}{r - 1}\right)^n$

Since LHS = RHS, the identity $P^2R^n = S^n$ is proven for the case $r \neq 1$.

Now consider the case when $r = 1$.

In this case, $S = na$, $P = a^n$, and $R = \frac{n}{a}$.

Left Hand Side (LHS) = $P^2R^n$

LHS = $(a^n)^2 \left(\frac{n}{a}\right)^n = a^{2n} \frac{n^n}{a^n}$

LHS = $a^{2n-n} n^n = a^n n^n = (an)^n = (na)^n$.

Right Hand Side (RHS) = $S^n$

Substitute the formula for S (when $r = 1$):

RHS = $(na)^n$.

Comparing the LHS and RHS for $r = 1$:

LHS = $(na)^n$

RHS = $(na)^n$

Since LHS = RHS, the identity $P^2R^n = S^n$ is proven for the case $r = 1$.

Since the identity $P^2R^n = S^n$ holds for both $r \neq 1$ and $r = 1$, the proof is complete.

Therefore, $P^2R^n = S^n$ is proven.

Given:

a, b, c, and d are in Geometric Progression (G.P.).

To Prove:

$(a^n + b^n)$, $(b^n + c^n)$, and $(c^n + d^n)$ are in G.P.

Solution:

Since a, b, c, and d are in G.P., there exists a common ratio, let's call it $r$, such that:

We need to show that the three terms $(a^n + b^n)$, $(b^n + c^n)$, and $(c^n + d^n)$ are in G.P. Three terms, say X, Y, and Z, are in G.P. if $Y^2 = XZ$.

Let the three terms be:

$X = a^n + b^n$

$Y = b^n + c^n$

$Z = c^n + d^n$

Now, express these terms using 'a' and 'r' by substituting the relations from (i), (ii), and (iii):

$X = a^n + (ar)^n = a^n + a^n r^n = a^n(1 + r^n)$

$Y = (ar)^n + (ar^2)^n = a^n r^n + a^n r^{2n} = a^n r^n(1 + r^n)$

$Z = (ar^2)^n + (ar^3)^n = a^n r^{2n} + a^n r^{3n} = a^n r^{2n}(1 + r^n)$

Now, let's calculate $Y^2$ and $XZ$ and check if they are equal.

Calculate $Y^2$:

$Y^2 = (a^n r^n(1 + r^n))^2$

$Y^2 = (a^n)^2 (r^n)^2 (1 + r^n)^2$

Calculate $XZ$:

$XZ = [a^n(1 + r^n)] \times [a^n r^{2n}(1 + r^n)]$

$XZ = a^n \cdot a^n \cdot r^{2n} \cdot (1 + r^n) \cdot (1 + r^n)$

Comparing equations (iv) and (v), we see that:

$Y^2 = XZ$

This holds true regardless of the values of $a$, $r$, and $n$ (provided the terms $a^n, b^n, c^n, d^n$ are well-defined). In the case where $1+r^n = 0$, we have $X=0, Y=0, Z=0$, and $0^2 = 0 \times 0$, which is true. If $a=0$, then $a=b=c=d=0$, and the terms are $0, 0, 0$, which is also a G.P.

Since $Y^2 = XZ$, the three terms $(a^n + b^n)$, $(b^n + c^n)$, and $(c^n + d^n)$ are in G.P.

Alternatively, we can check the ratio of consecutive terms (assuming the terms are non-zero):

Ratio of the second term to the first term:

$\frac{b^n + c^n}{a^n + b^n} = \frac{a^n r^n(1 + r^n)}{a^n(1 + r^n)}$

Assuming $a \neq 0$ and $1 + r^n \neq 0$:

$\frac{b^n + c^n}{a^n + b^n} = r^n$

Ratio of the third term to the second term:

$\frac{c^n + d^n}{b^n + c^n} = \frac{a^n r^{2n}(1 + r^n)}{a^n r^n(1 + r^n)}$

Assuming $a \neq 0$, $r \neq 0$, and $1 + r^n \neq 0$:

$\frac{c^n + d^n}{b^n + c^n} = \frac{r^{2n}}{r^n} = r^{n}$

Since the ratios of consecutive terms are equal (both are $r^n$), the terms $(a^n + b^n)$, $(b^n + c^n)$, and $(c^n + d^n)$ form a G.P. (provided the terms are non-zero. If the terms are zero, as shown with the $Y^2=XZ$ method, they also form a G.P.).

Conclusion:

Thus, it is proven that if a, b, c, d are in G.P., then $(a^n + b^n)$, $(b^n + c^n)$, and $(c^n + d^n)$ are also in G.P.

Given:

1. $a$ and $b$ are the roots of the quadratic equation $x^2 - 3x + p = 0$.

2. $c$ and $d$ are the roots of the quadratic equation $x^2 - 12x + q = 0$.

3. a, b, c, and d form a Geometric Progression (G.P.).

To Prove:

$(q + p) : (q - p) = 17 : 15$

Solution:

From Vieta's formulas for the roots of a quadratic equation $Ax^2 + Bx + C = 0$, the sum of the roots is $-\frac{B}{A}$ and the product of the roots is $\frac{C}{A}$.

For the equation $x^2 - 3x + p = 0$, the roots are $a$ and $b$. Thus:

For the equation $x^2 - 12x + q = 0$, the roots are $c$ and $d$. Thus:

Since a, b, c, d are in G.P., let the common ratio be $r$. Then we can write the terms as:

$b = ar$

$c = ar^2$

$d = ar^3$

Substitute these into equations (1) and (3):

From (1): $a + ar = 3 \implies a(1+r) = 3$

From (3): $ar^2 + ar^3 = 12 \implies ar^2(1+r) = 12$

Now, we can solve equations (5) and (6) for $a$ and $r$. Divide equation (6) by equation (5), assuming $a \neq 0$ and $1+r \neq 0$:

$\frac{ar^2(1+r)}{a(1+r)} = \frac{12}{3}$

$r^2 = 4$

This gives two possible values for $r$: $r = 2$ or $r = -2$. Note that if $r=-1$, equation (5) becomes $a(1-1)=3 \implies 0=3$, which is impossible. Thus, $r \neq -1$, and the division was valid.

Case 1: $r = 2$

Substitute $r = 2$ into equation (5):

$a(1 + 2) = 3$

$3a = 3$

$a = 1$

Now find $p$ and $q$ using equations (2) and (4) and the G.P. relations:

$p = ab = a(ar) = a^2r = (1)^2(2) = 1 \times 2 = 2$

$q = cd = (ar^2)(ar^3) = a^2r^5 = (1)^2(2)^5 = 1 \times 32 = 32$

So, for this case, $p = 2$ and $q = 32$.

Calculate the ratio $(q+p) : (q-p)$:

$q + p = 32 + 2 = 34$

$q - p = 32 - 2 = 30$

$(q + p) : (q - p) = 34 : 30$

Divide both parts by 2:

$34 \div 2 = 17$

$30 \div 2 = 15$

The ratio is $17 : 15$.

Case 2: $r = -2$

Substitute $r = -2$ into equation (5):

$a(1 + (-2)) = 3$

$a(-1) = 3$

$a = -3$

Now find $p$ and $q$ using equations (2) and (4) and the G.P. relations:

$p = a^2r = (-3)^2(-2) = 9 \times (-2) = -18$

$q = a^2r^5 = (-3)^2(-2)^5 = 9 \times (-32) = -288$

So, for this case, $p = -18$ and $q = -288$.

Calculate the ratio $(q+p) : (q-p)$:

$q + p = -288 + (-18) = -306$

$q - p = -288 - (-18) = -288 + 18 = -270$

$(q + p) : (q - p) = -306 : -270$

$\frac{-306}{-270} = \frac{306}{270}$

Simplify the fraction $\frac{306}{270}$ by dividing the numerator and denominator by their greatest common divisor. Both are divisible by 18:

$\frac{\cancel{306}^{17}}{\cancel{270}_{15}}$

The ratio is $17 : 15$.

In both possible cases for the common ratio $r$, the ratio $(q+p) : (q-p)$ is $17 : 15$.

Conclusion:

Therefore, if a, b, c, d are in G.P., then $(q + p) : (q - p) = 17 : 15$ is proven.

Given:

Two positive numbers are $a$ and $b$.

The Arithmetic Mean (A.M.) of $a$ and $b$ is $A = \frac{a+b}{2}$.

The Geometric Mean (G.M.) of $a$ and $b$ is $G = \sqrt{ab}$.

The ratio of the A.M. and G.M. is $m : n$, which means $\frac{A}{G} = \frac{m}{n}$.

To Show:

$a : b = \left( m+\sqrt{m^{2}-n^{2}} \right) : \left( m-\sqrt{m^{2}-n^{2}} \right)$

This is equivalent to showing that $\frac{a}{b} = \frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}$.

Solution:

We are given the ratio of A.M. to G.M.:

$\frac{A}{G} = \frac{m}{n}$

Substitute the formulas for A.M. and G.M.:

$\frac{\frac{a+b}{2}}{\sqrt{ab}} = \frac{m}{n}$

$\frac{a+b}{2\sqrt{ab}} = \frac{m}{n}$

We can rewrite this equality as:

$\frac{a+b}{m} = \frac{2\sqrt{ab}}{n}$

Let this common ratio be $k$ (assuming $m, n \neq 0$ and $a, b > 0$, so $k > 0$):

Square equation (2):

$(2\sqrt{ab})^2 = (nk)^2$

Consider the algebraic identity $(a-b)^2 = (a+b)^2 - 4ab$.

Substitute the expressions for $(a+b)$ from (1) and $4ab$ from (3) into this identity:

$(a-b)^2 = (mk)^2 - n^2k^2$

$(a-b)^2 = m^2k^2 - n^2k^2$

$(a-b)^2 = k^2(m^2 - n^2)$

Taking the square root of both sides (since $a$ and $b$ are interchangeable in the definition of AM and GM, we can assume $a \ge b$ without loss of generality for the ratio $a:b$. By AM-GM inequality, $\frac{a+b}{2} \ge \sqrt{ab}$, which implies $\frac{m}{n} \ge 1$, so $m \ge n$. Thus $m^2 - n^2 \ge 0$. Also $k>0$):

Now we have a system of two linear equations in $a$ and $b$:

$a+b = mk$ (from 1)

$a-b = k\sqrt{m^2 - n^2}$ (from 4)

Add the two equations to find $a$:

$(a+b) + (a-b) = mk + k\sqrt{m^2 - n^2}$

$2a = k(m + \sqrt{m^2 - n^2})$

Subtract the second equation from the first to find $b$:

$(a+b) - (a-b) = mk - k\sqrt{m^2 - n^2}$

$2b = k(m - \sqrt{m^2 - n^2})$

Now, find the ratio $a:b$ by dividing the expression for $2a$ by the expression for $2b$ (since $b > 0$, $2b \neq 0$. Also $m > n$ for $\sqrt{m^2-n^2} > 0$, if $m=n$, then $a=b$ and the formula holds as $1:1$):

$\frac{2a}{2b} = \frac{k(m + \sqrt{m^2 - n^2})}{k(m - \sqrt{m^2 - n^2})}$

Since $k > 0$, we can cancel $k$ from the numerator and denominator:

$\frac{a}{b} = \frac{m + \sqrt{m^2 - n^2}}{m - \sqrt{m^2 - n^2}}$

This can be expressed as the ratio $a : b$:

$a : b = (m + \sqrt{m^2 - n^2}) : (m - \sqrt{m^2 - n^2})$

Conclusion:

Thus, it is shown that $a : b = \left( m+\sqrt{m^{2}-n^{2}} \right) : \left( m-\sqrt{m^{2}-n^{2}} \right)$.

We need to find the sum of the given series up to n terms.

(i) 5 + 55 + 555 + …

Let $S_n$ be the sum of the first n terms of the series.

$S_n = 5 + 55 + 555 + \dots + \text{up to n terms}$

Factor out 5:

$S_n = 5(1 + 11 + 111 + \dots + \text{up to n terms})$

Multiply and divide by 9:

$S_n = \frac{5}{9}(9 + 99 + 999 + \dots + \text{up to n terms})$

Rewrite each term inside the bracket using powers of 10:

$S_n = \frac{5}{9}((10-1) + (10^2-1) + (10^3-1) + \dots + (10^n-1))$

Group the positive terms and the negative terms:

$S_n = \frac{5}{9}[(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + \text{n terms})]$

The first bracket contains a Geometric Progression (G.P.) with first term $a=10$, common ratio $r=10$, and number of terms $n$. The sum of this G.P. is $\frac{a(r^n - 1)}{r - 1}$.

Sum of the G.P. $= \frac{10(10^n - 1)}{10 - 1} = \frac{10}{9}(10^n - 1)$.

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

Substitute these sums back into the expression for $S_n$:

$S_n = \frac{5}{9}\left[\frac{10}{9}(10^n - 1) - n\right]$

Distribute the $\frac{5}{9}$:

$S_n = \frac{5}{9} \times \frac{10}{9}(10^n - 1) - \frac{5}{9}n$

$S_n = \frac{50}{81}(10^n - 1) - \frac{5n}{9}$

(ii) .6 + .66 + .666 + …

Let $S'_n$ be the sum of the first n terms of the series.

$S'_n = 0.6 + 0.66 + 0.666 + \dots + \text{up to n terms}$

Factor out 6:

$S'_n = 6(0.1 + 0.11 + 0.111 + \dots + \text{up to n terms})$

Multiply and divide by 9:

$S'_n = \frac{6}{9}(0.9 + 0.99 + 0.999 + \dots + \text{up to n terms})$

Simplify $\frac{6}{9}$ to $\frac{2}{3}$ and rewrite each term inside the bracket:

$S'_n = \frac{2}{3}((1 - 0.1) + (1 - 0.01) + (1 - 0.001) + \dots + (1 - 10^{-n}))$

Group the positive terms and the negative terms:

$S'_n = \frac{2}{3}[(1 + 1 + 1 + \dots + \text{n terms}) - (0.1 + 0.01 + 0.001 + \dots + 10^{-n})]$

The first bracket contains the sum of n ones, which is $n$.

The second bracket contains a Geometric Progression (G.P.) with first term $a=0.1 = \frac{1}{10}$, common ratio $r=0.1 = \frac{1}{10}$, and number of terms $n$. The sum of this G.P. is $\frac{a(1 - r^n)}{1 - r}$.

Sum of the G.P. $= \frac{\frac{1}{10}(1 - (\frac{1}{10})^n)}{1 - \frac{1}{10}} = \frac{\frac{1}{10}(1 - 10^{-n})}{\frac{9}{10}} = \frac{1}{10} \times \frac{10}{9} (1 - 10^{-n}) = \frac{1}{9}(1 - 10^{-n})$.

Substitute these sums back into the expression for $S'_n$:

$S'_n = \frac{2}{3}\left[n - \frac{1}{9}(1 - 10^{-n})\right]$

Distribute the $\frac{2}{3}$:

$S'_n = \frac{2}{3}n - \frac{2}{3} \times \frac{1}{9}(1 - 10^{-n})$

$S'_n = \frac{2n}{3} - \frac{2}{27}(1 - 10^{-n})$

We are asked to find the 20^th term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots$

First, let's analyze the pattern of the terms in the series.

The first term is $T_1 = 2 \times 4$.

The second term is $T_2 = 4 \times 6$.

The third term is $T_3 = 6 \times 8$.

We observe that each term is a product of two numbers.

Let's look at the first factors in each term: 2, 4, 6, ...

This sequence is an Arithmetic Progression (A.P.) with the first term $a_1 = 2$ and the common difference $d = 4 - 2 = 2$.

The n-th term of this A.P. is given by the formula $a_n = a_1 + (n-1)d$.

$a_n = 2 + (n-1)2 = 2 + 2n - 2 = 2n$.

So, the first factor of the n-th term of the series is $2n$.

Now, let's look at the second factors in each term: 4, 6, 8, ...

This sequence is also an Arithmetic Progression (A.P.) with the first term $b_1 = 4$ and the common difference $d = 6 - 4 = 2$.

The n-th term of this A.P. is given by the formula $b_n = b_1 + (n-1)d$.

$b_n = 4 + (n-1)2 = 4 + 2n - 2 = 2n + 2$.

So, the second factor of the n-th term of the series is $2n + 2$.

The n-th term of the given series, $T_n$, is the product of the n-th terms of these two A.P.s:

$T_n = (2n)(2n + 2)$

$T_n = 4n(n + 1)$

$T_n = 4n^2 + 4n$

We need to find the 20^th term of the series. To do this, we substitute $n = 20$ into the formula for $T_n$:

$T_{20} = 4(20)(20 + 1)$

$T_{20} = 4(20)(21)$

$T_{20} = 80 \times 21$

Now, perform the multiplication:

$80 \times 21 = 80 \times (20 + 1) = 80 \times 20 + 80 \times 1 = 1600 + 80 = 1680$.

So, $T_{20} = 1680$.

The 20^th term of the series is $\mathbf{1680}$.

Given:

Original price of the tractor = $\textsf{₹}12000$

Cash paid = $\textsf{₹}6000$

Annual principal instalment = $\textsf{₹}500$

Interest rate = 12% per annum on the unpaid amount.

To Find:

The total cost of the tractor.

Solution:

First, calculate the balance amount to be paid through instalments.

Balance Amount = Original Price - Cash Paid

Balance Amount = $\textsf{₹}12000 - \textsf{₹}6000 = \textsf{₹}6000$

This balance amount is the principal loan that needs to be repaid.

The principal is repaid in annual instalments of $\textsf{₹}500$. To find the number of instalments, divide the total principal balance by the annual principal payment.

Number of instalments = $\frac{\text{Total Principal Balance}}{\text{Annual Principal Payment}}$

Number of instalments = $\frac{\textsf{₹}6000}{\textsf{₹}500} = 12$

There will be 12 annual instalments.

Interest is paid at 12% on the unpaid amount each year. The unpaid amount decreases by $\textsf{₹}500$ each year.

The unpaid amounts at the beginning of each year (on which interest is calculated) are:

Year 1: $\textsf{₹}6000$

Year 2: $\textsf{₹}6000 - \textsf{₹}500 = \textsf{₹}5500$

Year 3: $\textsf{₹}5500 - \textsf{₹}500 = \textsf{₹}5000$

... and so on, until the last payment.

Year 12: $\textsf{₹}500$ (This is the remaining principal before the 12th principal payment)

The sequence of unpaid amounts on which interest is calculated is an Arithmetic Progression:

$6000, 5500, 5000, \dots, 500$

This A.P. has the first term $a_1 = 6000$, the common difference $d = -500$, and the number of terms $n = 12$.

The total interest paid is 12% of the sum of these unpaid amounts over the 12 years.

First, calculate the sum of this A.P. using the formula $S_n = \frac{n}{2}(a_1 + a_n)$.

Sum of unpaid amounts $= \frac{12}{2}(6000 + 500)$

Sum of unpaid amounts $= 6(6500) = 39000$

The total amount on which interest is charged over the 12 years is $\textsf{₹}39000$.

Now, calculate the total interest paid at 12%.

Total Interest Paid = 12% of $\textsf{₹}39000$

Total Interest Paid $= \frac{12}{100} \times 39000$

Total Interest Paid $= 0.12 \times 39000 = 4680$

The total interest paid is $\textsf{₹}4680$.

The total cost of the tractor to the farmer is the sum of the initial cash payment, the total principal repaid, and the total interest paid.

Total Principal Repaid = Number of instalments $\times$ Annual Principal Payment

Total Principal Repaid = $12 \times \textsf{₹}500 = \textsf{₹}6000$ (This is the balance amount that was loaned)

Total Cost = Cash Paid + Total Principal Repaid + Total Interest Paid

Total Cost = $\textsf{₹}6000 + \textsf{₹}6000 + \textsf{₹}4680$

Total Cost = $\textsf{₹}12000 + \textsf{₹}4680 = \textsf{₹}16680$

Answer:

The total cost of the tractor to the farmer is $\textsf{₹}16680$.

Given:

Original price of the scooter = $\textsf{₹}22000$

Cash paid = $\textsf{₹}4000$

Annual principal instalment = $\textsf{₹}1000$

Interest rate = 10% per annum on the unpaid amount.

To Find:

The total cost of the scooter.

Solution:

First, calculate the balance amount to be paid through instalments.

Balance Amount = Original Price - Cash Paid

Balance Amount = $\textsf{₹}22000 - \textsf{₹}4000 = \textsf{₹}18000$

This balance amount is the principal loan that needs to be repaid.

The principal is repaid in annual instalments of $\textsf{₹}1000$. To find the number of instalments, divide the total principal balance by the annual principal payment.

Number of instalments = $\frac{\text{Total Principal Balance}}{\text{Annual Principal Payment}}$

Number of instalments = $\frac{\textsf{₹}18000}{\textsf{₹}1000} = 18$

There will be 18 annual instalments.

Interest is paid at 10% on the unpaid amount each year. The unpaid amount decreases by $\textsf{₹}1000$ each year.

The unpaid amounts at the beginning of each year (on which interest is calculated) are:

Year 1: $\textsf{₹}18000$

Year 2: $\textsf{₹}18000 - \textsf{₹}1000 = \textsf{₹}17000$

Year 3: $\textsf{₹}17000 - \textsf{₹}1000 = \textsf{₹}16000$

... and so on, until the last payment.

Year 18: $\textsf{₹}1000$ (This is the remaining principal before the 18th principal payment)

The sequence of unpaid amounts on which interest is calculated is an Arithmetic Progression (A.P.):

$18000, 17000, 16000, \dots, 1000$

This A.P. has the first term $a_1 = 18000$, the common difference $d = -1000$, and the number of terms $n = 18$. The last term is $a_{18} = 1000$.

The total interest paid is 10% of the sum of these unpaid amounts over the 18 years.

First, calculate the sum of this A.P. using the formula for the sum of an A.P., $S_n = \frac{n}{2}(a_1 + a_n)$.

Sum of unpaid amounts $= \frac{18}{2}(18000 + 1000)$

Sum of unpaid amounts $= 9(19000) = 171000$

The total amount on which interest is charged over the 18 years is $\textsf{₹}171000$.

Now, calculate the total interest paid at 10%.

Total Interest Paid = 10% of $\textsf{₹}171000$

Total Interest Paid $= \frac{10}{100} \times 171000$

Total Interest Paid $= 0.10 \times 171000 = 17100$

The total interest paid is $\textsf{₹}17100$.

The total cost of the scooter to Shamshad Ali is the sum of the initial cash payment, the total principal repaid, and the total interest paid.

Total Principal Repaid = Number of instalments $\times$ Annual Principal Payment

Total Principal Repaid = $18 \times \textsf{₹}1000 = \textsf{₹}18000$ (This is the balance amount that was loaned)

Total Cost = Cash Paid + Total Principal Repaid + Total Interest Paid

Total Cost = $\textsf{₹}4000 + \textsf{₹}18000 + \textsf{₹}17100$

Total Cost = $\textsf{₹}22000 + \textsf{₹}17100 = \textsf{₹}39100$

Answer:

The total cost of the scooter to Shamshad Ali is $\textsf{₹}39100$.

Given:

Initial number of letters sent = 4

Each person who receives a letter sends it to 4 different persons.

Cost to mail one letter = 50 paise = $\textsf{₹}0.50$

To Find:

The total amount spent on postage when the 8^th set of letters is mailed.

Solution:

Let the number of letters mailed in each set form a sequence.

In the 1^st set, the person writes to 4 friends. Number of letters = 4.

In the 2^nd set, each of the 4 friends writes to 4 persons. Number of letters = $4 \times 4 = 16$.

In the 3^rd set, each of the 16 recipients from the 2^nd set writes to 4 persons. Number of letters = $16 \times 4 = 64$.

This sequence of the number of letters mailed in each set is a Geometric Progression (G.P.):

$4, 16, 64, \dots$

The first term of this G.P. is $a = 4$.

The common ratio is $r = \frac{16}{4} = 4$.

We need to find the total amount spent when the 8^th set of letters is mailed. This means we need the total number of letters mailed from the 1^st set up to the 8^th set.

The total number of letters is the sum of the first 8 terms of this G.P.

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

We need to find the sum of the first 8 terms, so $n = 8$.

$S_8 = \frac{4(4^8 - 1)}{4 - 1}$

$S_8 = \frac{4(4^8 - 1)}{3}$

Calculate $4^8$:

$4^1 = 4$

$4^2 = 16$

$4^3 = 64$

$4^4 = 256$

$4^5 = 1024$

$4^6 = 4096$

$4^7 = 16384$

$4^8 = 65536$

Substitute the value of $4^8$ into the formula for $S_8$:

$S_8 = \frac{4(65536 - 1)}{3}$

$S_8 = \frac{4(65535)}{3}$

Divide 65535 by 3:

$\frac{65535}{3} = 21845$

$S_8 = 4 \times 21845$

$S_8 = 87380$

The total number of letters mailed up to the 8^th set is 87380.

The cost to mail one letter is 50 paise, which is equivalent to $\textsf{₹}0.50$.

The total amount spent on postage is the total number of letters multiplied by the cost per letter.

Total Cost = Total Number of Letters $\times$ Cost per Letter

Total Cost = $87380 \times \textsf{₹}0.50$

Total Cost = $87380 \times \frac{1}{2}$

Total Cost = $\frac{87380}{2}$

Total Cost = $\textsf{₹}43690$

Answer:

The amount spent on the postage when the 8^th set of letters is mailed is $\textsf{₹}43690$.

Given:

Principal amount deposited ($P$) = $\textsf{₹}10000$

Annual Simple Interest Rate ($R$) = 5%

To Find:

1. The amount in the 15^th year (at the end of 15 years).

2. The total amount after 20 years (at the end of 20 years).

Solution:

For simple interest, the interest earned each year is constant and is calculated only on the principal amount.

The formula for simple interest ($I$) for $T$ years is given by:

$I = \frac{P \times R \times T}{100}$

The total amount ($A$) after $T$ years is given by:

$A = P + I = P + \frac{P \times R \times T}{100} = P\left(1 + \frac{RT}{100}\right)$

First, let's calculate the interest earned per year:

Annual Interest $= \frac{10000 \times 5 \times 1}{100} = \frac{50000}{100} = \textsf{₹}500$

So, the interest earned each year is $\textsf{₹}500$.

Now, let's find the amount in the 15^th year (which means at the end of 15 years).

Here, the time period is $T = 15$ years.

Total Interest after 15 years $= \textsf{₹}500 \times 15 = \textsf{₹}7500$

Amount at the end of 15 years $= \textsf{₹}10000$ (Principal) + $\textsf{₹}7500$ (Total Interest)

Amount at the end of 15 years $= \textsf{₹}17500$

Next, let's calculate the total amount after 20 years (at the end of 20 years).

Here, the time period is $T = 20$ years.

Total Interest after 20 years $= \textsf{₹}500 \times 20 = \textsf{₹}10000$

Amount at the end of 20 years $= \textsf{₹}10000$ (Principal) + $\textsf{₹}10000$ (Total Interest)

Amount at the end of 20 years $= \textsf{₹}20000$

Answer:

The amount in the 15^th year (at the end of 15 years) is $\mathbf{\textsf{₹}17500}$.

The total amount after 20 years is $\mathbf{\textsf{₹}20000}$.

Given:

Cost of the machine (Initial Value, $V_0$) = $\textsf{₹}15625$

Annual Depreciation Rate ($r$) = 20%

Time period ($t$) = 5 years

To Find:

Estimated value of the machine at the end of 5 years ($V_5$).

Solution:

When an asset depreciates by a fixed percentage each year, its value decreases in a pattern similar to how an investment grows with compound interest, but using a subtraction instead of addition in the formula.

The formula for the value of an asset ($V_t$) after $t$ years, with an initial value $V_0$ and an annual depreciation rate $r$ (as a percentage), is:

$V_t = V_0 \left(1 - \frac{r}{100}\right)^t$

In this problem, $V_0 = 15625$, $r = 20$, and $t = 5$. We need to find $V_5$.

First, calculate the depreciation factor $(1 - \frac{r}{100})$:

$1 - \frac{20}{100} = 1 - \frac{1}{5} = \frac{5 - 1}{5} = \frac{4}{5}$

Now, substitute the values into the formula:

$V_5 = 15625 \left(\frac{4}{5}\right)^5$

$V_5 = 15625 \times \frac{4^5}{5^5}$

Calculate $4^5$ and $5^5$:

$4^5 = 4 \times 4 \times 4 \times 4 \times 4 = 16 \times 16 \times 4 = 256 \times 4 = 1024$

$5^5 = 5 \times 5 \times 5 \times 5 \times 5 = 25 \times 25 \times 5 = 625 \times 5 = 3125$

Substitute these values back into the expression for $V_5$:

$V_5 = 15625 \times \frac{1024}{3125}$

We can simplify this by dividing 15625 by 3125.

Let's check how many times 3125 goes into 15625:

$15625 \div 3125 = 5$ (Since $5 \times 3125 = 15625$)

So, we can cancel the terms:

$V_5 = \cancel{15625}^{5} \times \frac{1024}{\cancel{3125}_{1}}$

$V_5 = 5 \times 1024$

$V_5 = 5120$

So, the estimated value of the machine at the end of 5 years is $\textsf{₹}5120$.

Answer:

The estimated value of the machine at the end of 5 years is $\textsf{₹}5120$.

Given:

Initial number of workers = 150

Number of workers dropping out each day (starting from the second day) = 4

Extra days taken to finish the work = 8

To Find:

The total number of days in which the work was completed.

Solution:

Let $N$ be the original number of days planned to complete the work with 150 workers.

Assume that each worker does one unit of work per day.

The total amount of work to be done is the product of the number of workers and the original number of days.

Due to workers dropping out, the number of workers on successive days forms an Arithmetic Progression (A.P.).

Number of workers on Day 1 = 150

Number of workers on Day 2 = $150 - 4 = 146$

Number of workers on Day 3 = $146 - 4 = 142$

... and so on.

This is an A.P. with the first term $a_1 = 150$ and a common difference $d = -4$.

The work took 8 more days than planned, so the actual number of days taken to complete the work is $N+8$.

The total work done is the sum of the number of workers present on each day for $N+8$ days.

The total work done is the sum of the first $N+8$ terms of the A.P. representing the number of workers each day.

The sum of the first $k$ terms of an A.P. is given by $S_k = \frac{k}{2}[2a_1 + (k-1)d]$.

Here, $k = N+8$, $a_1 = 150$, and $d = -4$.

Total Work Done = $S_{N+8} = \frac{N+8}{2}[2(150) + (N+8-1)(-4)]$

$S_{N+8} = \frac{N+8}{2}[300 + (N+7)(-4)]$

$S_{N+8} = \frac{N+8}{2}[300 - 4N - 28]$

$S_{N+8} = \frac{N+8}{2}[272 - 4N]$

Factor out 2 from the bracket:

$S_{N+8} = \frac{N+8}{\cancel{2}}\cancel{(2)}[136 - 2N]$

The total work is the same regardless of how it was completed. Therefore, we equate the Total Work from (1) and (2):

$150N = (N+8)(136 - 2N)$

$150N = 136N - 2N^2 + 8 \times 136 - 8 \times 2N$

$150N = 136N - 2N^2 + 1088 - 16N$

Combine like terms on the right side:

$150N = (136N - 16N) - 2N^2 + 1088$

$150N = 120N - 2N^2 + 1088$

Move all terms to one side to form a quadratic equation:

$2N^2 + 150N - 120N - 1088 = 0$

$2N^2 + 30N - 1088 = 0$

Divide the entire equation by 2:

Now, we solve the quadratic equation for $N$. We can use factorization or the quadratic formula. Let's use the quadratic formula $N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=15$, $c=-544$.

$N = \frac{-15 \pm \sqrt{15^2 - 4(1)(-544)}}{2(1)}$

$N = \frac{-15 \pm \sqrt{225 + 2176}}{2}$

$N = \frac{-15 \pm \sqrt{2401}}{2}$

Find the square root of 2401. $\sqrt{2401} = 49$.

$N = \frac{-15 \pm 49}{2}$

We get two possible values for $N$:

$N_1 = \frac{-15 + 49}{2} = \frac{34}{2} = 17$

$N_2 = \frac{-15 - 49}{2} = \frac{-64}{2} = -32$

Since the number of days must be positive, we take $N = 17$.

The original number of days planned was 17.

The work was completed in 8 more days than planned.

Number of days in which the work was completed = $N + 8 = 17 + 8 = 25$ days.

We should verify that the number of workers remains positive throughout the 25 days. The number of workers on day $k$ is $154 - 4k$. On day 25, the number of workers is $154 - 4(25) = 154 - 100 = 54$, which is positive.

Answer:

The number of days in which the work was completed is $\mathbf{25}$.