Chapter 9 Sequences and Series (Additional Questions)

Given:

The given Arithmetic Progression (AP) is 2, 5, 8, 11, ...

To Find:

The 15th term of the given AP.

Solution:

The given AP is 2, 5, 8, 11, ...

The first term is $a = 2$.

The common difference is $d = 5 - 2 = 3$.

Alternatively, $d = 8 - 5 = 3$, and so on.

The formula for the $n$th term of an AP is given by:

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

We need to find the 15th term, so we set $n = 15$.

Substituting the values of $a$, $d$, and $n$ into the formula, we get:

$a_{15} = 2 + (15-1) \times 3$

$a_{15} = 2 + (14) \times 3$

$a_{15} = 2 + 42$

$a_{15} = 44$

Therefore, the 15th term of the given AP is 44.

The correct option is (A) 44.

Given:

The given Arithmetic Progression (AP) is 1, 3, 5, 7, ...

We need to find the sum of the first 20 terms, so $n=20$.

To Find:

The sum of the first 20 terms ($S_{20}$) of the given AP.

Solution:

The given AP is 1, 3, 5, 7, ...

The first term is $a = 1$.

The common difference is $d = 3 - 1 = 2$.

Alternatively, $d = 5 - 3 = 2$, and so on.

The number of terms is $n = 20$.

The formula for the sum of the first $n$ terms of an AP is given by:

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

We need to find the sum of the first 20 terms, so we set $n = 20$.

Substituting the values of $a$, $d$, and $n$ into the formula, we get:

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

$S_{20} = 10[2 + (19)2]$

$S_{20} = 10[2 + 38]$

$S_{20} = 10[40]$

$S_{20} = 400$

Therefore, the sum of the first 20 terms of the given AP is 400.

The correct option is (A) 400.

Given:

First term of the AP, $a = 5$.

Common difference, $d = 3$.

The $n^{th}$ term of the AP, $a_n = 50$.

To Find:

The value of $n$.

Solution:

The formula for the $n^{th}$ term of an AP is given by:

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

We are given $a = 5$, $d = 3$, and $a_n = 50$. Substituting these values into the formula, we get:

$50 = 5 + (n-1)3$

Subtract 5 from both sides:

$50 - 5 = (n-1)3$

$45 = (n-1)3$

Divide both sides by 3:

$\frac{45}{3} = n-1$

$15 = n-1$

Add 1 to both sides:

$15 + 1 = n$

$n = 16$

Therefore, the value of $n$ is 16.

The correct option is (B) 16.

Given:

The two numbers are 10 and 28.

We need to insert two arithmetic means between them.

To Find:

The two arithmetic means between 10 and 28.

Solution:

Let the two arithmetic means between 10 and 28 be $A_1$ and $A_2$.

Then the sequence 10, $A_1$, $A_2$, 28 forms an Arithmetic Progression (AP).

In this AP:

The first term is $a = 10$.

The last term (which is the 4th term) is $a_4 = 28$.

The number of terms is $n = 4$.

Let $d$ be the common difference of the AP.

The formula for the $n^{th}$ term of an AP is $a_n = a + (n-1)d$.

Using this formula for the 4th term:

$a_4 = a + (4-1)d$

$28 = 10 + 3d$

Subtracting 10 from both sides:

$28 - 10 = 3d$

$18 = 3d$

Dividing by 3:

$d = \frac{18}{3}$

$d = 6$

Now we can find the arithmetic means $A_1$ and $A_2$.

$A_1$ is the 2nd term of the AP:

$A_1 = a + d$

$A_1 = 10 + 6$

$A_1 = 16$

$A_2$ is the 3rd term of the AP:

$A_2 = a + 2d$

$A_2 = 10 + 2(6)$

$A_2 = 10 + 12$

$A_2 = 22$

The two arithmetic means between 10 and 28 are 16 and 22.

The resulting AP is 10, 16, 22, 28.

The correct option is (A) 16, 22.

Given:

The given Geometric Progression (GP) is 3, 6, 12, 24, ...

To Find:

The common ratio of the given GP.

Solution:

The common ratio ($r$) of a Geometric Progression is the ratio of any term to its preceding term.

The given GP is 3, 6, 12, 24, ...

The first term is $a_1 = 3$.

The second term is $a_2 = 6$.

The third term is $a_3 = 12$.

The common ratio $r$ can be calculated as:

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

$r = \frac{6}{3}$

$r = 2$

We can verify this with other terms:

$r = \frac{a_3}{a_2} = \frac{12}{6} = 2$

$r = \frac{a_4}{a_3} = \frac{24}{12} = 2$

The common ratio of the given GP is 2.

The correct option is (A) 2.

Given:

The given Geometric Progression (GP) is 2, 6, 18, ...

We need to find the $7^{th}$ term, so $n=7$.

To Find:

The $7^{th}$ term ($a_7$) of the given GP.

Solution:

The given GP is 2, 6, 18, ...

The first term is $a = 2$.

The common ratio is $r = \frac{\text{second term}}{\text{first term}} = \frac{6}{2} = 3$.

Alternatively, $r = \frac{\text{third term}}{\text{second term}} = \frac{18}{6} = 3$.

The formula for the $n^{th}$ term of a GP is given by:

$a_n = ar^{n-1}$

We need to find the $7^{th}$ term, so we set $n = 7$.

Substituting the values of $a$, $r$, and $n$ into the formula, we get:

$a_7 = 2 \times (3)^{7-1}$

$a_7 = 2 \times (3)^6$

Calculate $3^6$:

$3^1 = 3$

$3^2 = 9$

$3^3 = 27$

$3^4 = 81$

$3^5 = 243$

$3^6 = 729$

Now, substitute the value of $3^6$ back into the formula for $a_7$:

$a_7 = 2 \times 729$

$a_7 = 1458$

Therefore, the $7^{th}$ term of the given GP is 1458.

The correct option is (A) 1458.

Given:

The given Geometric Progression (GP) is 1, 2, 4, 8, ...

We need to find the sum of the first 6 terms, so $n=6$.

To Find:

The sum of the first 6 terms ($S_6$) of the given GP.

Solution:

The given GP is 1, 2, 4, 8, ...

The first term is $a = 1$.

The common ratio is $r = \frac{\text{second term}}{\text{first term}} = \frac{2}{1} = 2$.

Alternatively, $r = \frac{\text{third term}}{\text{second term}} = \frac{4}{2} = 2$, and so on.

The number of terms is $n = 6$.

Since the common ratio $r = 2$ is not equal to 1, the formula for the sum of the first $n$ terms of a GP is given by:

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

We need to find the sum of the first 6 terms, so we set $n = 6$.

Substituting the values of $a$, $r$, and $n$ into the formula, we get:

$S_6 = \frac{1(2^6 - 1)}{2 - 1}$

$S_6 = \frac{2^6 - 1}{1}$

$S_6 = 2^6 - 1$

Calculate $2^6$:

$2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$

Now, substitute the value of $2^6$ back into the formula for $S_6$:

$S_6 = 64 - 1$

$S_6 = 63$

Therefore, the sum of the first 6 terms of the given GP is 63.

The correct option is (B) 63.

Assertion (A): If $a, b, c$ are in AP, then $b = \frac{a+c}{2}$.

Reason (R): The arithmetic mean of two numbers $a$ and $c$ is $\frac{a+c}{2}$.

Analysis of Assertion (A):

If $a, b, c$ are in an Arithmetic Progression (AP), the difference between consecutive terms is constant. This common difference ($d$) is given by:

$b - a = d$

$c - b = d$

Therefore, we have:

$b - a = c - b$

Adding $b$ to both sides:

$b + b - a = c$

$2b - a = c$

Adding $a$ to both sides:

$2b = a + c$

Dividing by 2:

$b = \frac{a+c}{2}$

This shows that if $a, b, c$ are in AP, then $b$ is indeed the arithmetic mean of $a$ and $c$. Thus, Assertion (A) is true.

Analysis of Reason (R):

The arithmetic mean of two numbers is defined as their sum divided by 2. For two numbers $a$ and $c$, their arithmetic mean is $\frac{a+c}{2}$. This is a fundamental definition in arithmetic.

Thus, Reason (R) is true.

Relationship between Assertion (A) and Reason (R):

Assertion (A) states a property ($b = \frac{a+c}{2}$) that holds true if $a, b, c$ are in AP. Reason (R) provides the definition of the arithmetic mean. The property stated in Assertion (A) is exactly that the middle term $b$ is the arithmetic mean of the other two terms $a$ and $c$. Therefore, the definition provided in Reason (R) is the correct explanation for why Assertion (A) is true.

Conclusion:

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Given:

Types of progressions and their formulas/definitions.

To Match:

Match the progression type with the correct formula or definition.

Solution:

Let's analyze each type of progression and its general term or definition:

(i) Arithmetic Progression (AP):

An AP is a sequence where the difference between consecutive terms is constant. The general term (or $n^{th}$ term) of an AP is given by the formula:

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

where $a$ is the first term and $d$ is the common difference.

This matches option (a) $a + (n-1)d$.

So, (i) matches with (a).

(ii) Geometric Progression (GP):

A GP is a sequence where the ratio of any term to its preceding term is constant. This constant ratio is called the common ratio ($r$). The general term (or $n^{th}$ term) of a GP is given by the formula:

$a_n = ar^{n-1}$

where $a$ is the first term and $r$ is the common ratio.

This matches option (c) $ar^{n-1}$.

So, (ii) matches with (c).

(iii) Harmonic Progression (HP):

A Harmonic Progression (HP) is a sequence of non-zero numbers such that the reciprocals of the numbers form an Arithmetic Progression (AP).

This matches option (b) Reciprocals form an AP.

So, (iii) matches with (b).

Matching Summary:

(i) AP $\rightarrow$ (a) $a + (n-1)d$

(ii) GP $\rightarrow$ (c) $ar^{n-1}$

(iii) HP $\rightarrow$ (b) Reciprocals form an AP

Comparing this summary with the given options:

(A) (i)-(a), (ii)-(b), (iii)-(c) - Incorrect

(B) (i)-(a), (ii)-(c), (iii)-(b) - Correct

(C) (i)-(b), (ii)-(a), (iii)-(c) - Incorrect

(D) (i)-(c), (ii)-(a), (iii)-(b) - Incorrect

The correct matching is (i)-(a), (ii)-(c), (iii)-(b).

The correct option is (B).

Given:

An infinite Geometric Progression (GP) with first term $a$ and common ratio $r$.

The condition for the sum to exist is $|r| < 1$.

To Find:

The formula for the sum of this infinite GP.

Solution:

The sum of the first $n$ terms of a GP with first term $a$ and common ratio $r$ is given by:

$S_n = \frac{a(1 - r^n)}{1 - r}$, for $r \neq 1$.

For an infinite GP, we are interested in the sum as the number of terms $n$ approaches infinity, i.e., $S_\infty = \lim\limits_{n \to \infty} S_n$.

If $|r| < 1$, as $n$ approaches infinity, $r^n$ approaches 0 (i.e., $\lim\limits_{n \to \infty} r^n = 0$ when $|r| < 1$).

Substituting this into the formula for $S_n$:

$S_\infty = \lim\limits_{n \to \infty} \frac{a(1 - r^n)}{1 - r}$

$S_\infty = \frac{a(1 - \lim\limits_{n \to \infty} r^n)}{1 - r}$

$S_\infty = \frac{a(1 - 0)}{1 - r}$

$S_\infty = \frac{a}{1 - r}$

Thus, the sum of an infinite GP with first term $a$ and common ratio $r$ ($|r| < 1$) is $\frac{a}{1-r}$.

The correct option is (A) $\frac{a}{1-r}$.

Given:

The two numbers are 8 and 216.

We need to insert two geometric means between them.

To Find:

The two geometric means between 8 and 216.

Solution:

Let the two geometric means between 8 and 216 be $G_1$ and $G_2$.

Then the sequence 8, $G_1$, $G_2$, 216 forms a Geometric Progression (GP).

In this GP:

The first term is $a_1 = 8$.

The last term (which is the 4th term) is $a_4 = 216$.

The number of terms is $n = 4$.

Let $r$ be the common ratio of the GP.

The formula for the $n^{th}$ term of a GP is $a_n = a_1 r^{n-1}$.

Using this formula for the 4th term:

$a_4 = a_1 r^{4-1}$

$216 = 8 r^3$

Divide both sides by 8:

$r^3 = \frac{216}{8}$

$r^3 = 27$

Taking the cube root of both sides:

$r = \sqrt[3]{27}$

$r = 3$

Now we can find the geometric means $G_1$ and $G_2$.

$G_1$ is the 2nd term ($a_2$) of the GP:

$G_1 = a_1 r$

$G_1 = 8 \times 3$

$G_1 = 24$

$G_2$ is the 3rd term ($a_3$) of the GP:

$G_2 = a_1 r^2$

$G_2 = 8 \times 3^2$

$G_2 = 8 \times 9$

$G_2 = 72$

The two geometric means between 8 and 216 are 24 and 72.

The resulting GP is 8, 24, 72, 216.

The correct option is (A) 24, 72.

Given:

The arithmetic mean (AM) of two positive numbers $a$ and $b$ is 10.

The geometric mean (GM) of the same two positive numbers $a$ and $b$ is 8.

To Find:

The values of the two positive numbers $a$ and $b$.

Solution:

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

The formula for the arithmetic mean (AM) of $a$ and $b$ is $\frac{a+b}{2}$.

We are given that AM = 10.

The formula for the geometric mean (GM) of two positive numbers $a$ and $b$ is $\sqrt{ab}$.

We are given that GM = 8.

From equation (1), multiply both sides by 2:

From equation (2), square both sides:

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

From equation (3), we can express $b$ in terms of $a$:

$b = 20 - a$

Substitute this expression for $b$ into equation (4):

$a(20 - a) = 64$

Distribute $a$ on the left side:

$20a - a^2 = 64$

Rearrange the terms to form a quadratic equation:

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

We can solve this quadratic equation by factoring. We need two numbers that multiply to 64 and add up to -20. These numbers are -4 and -16.

So, the equation can be factored as:

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

This gives two possible values for $a$:

$a - 4 = 0 \implies a = 4$

or

$a - 16 = 0 \implies a = 16$

Now, we find the corresponding values for $b$ using $b = 20 - a$ (from equation 3):

If $a = 4$:

$b = 20 - 4 = 16$

If $a = 16$:

$b = 20 - 16 = 4$

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

Both numbers are positive, as required.

Let's verify the AM and GM for the pair (4, 16):

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

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

The two numbers are 4 and 16.

The correct option is (A) 4, 16.

Understanding the terms:

A sequence is an ordered list of numbers. The numbers in the list are called terms of the sequence. Example: $a_1, a_2, a_3, ...$

A series is the sum of the terms of a sequence. Example: $a_1 + a_2 + a_3 + ...$

Analyzing the options:

(A) $1 + 2 + 3 + ...$

This represents the sum of the terms of the sequence 1, 2, 3, ..., which is an arithmetic progression. This is explicitly written as a sum, so it is a series.

(B) $1, 4, 9, 16, ...$

This is an ordered list of numbers (the squares of natural numbers: $1^2, 2^2, 3^2, 4^2, ...$). The terms are separated by commas, indicating a list of numbers. This is a sequence.

It could be related to a series if we were to sum its terms (e.g., $1 + 4 + 9 + 16 + ...$), but the notation given ($1, 4, 9, 16, ...$) represents the list of terms itself, not their sum.

(C) $\sum\limits_{n=1}^{\infty} \frac{1}{n}$

The symbol $\sum$ (sigma) denotes summation. This expression represents the sum of the terms of the sequence $\frac{1}{n}$ for $n = 1, 2, 3, ...$. The sequence is $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ...$. The expression $\sum\limits_{n=1}^{\infty} \frac{1}{n}$ represents the infinite sum $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ...$. This is a series (specifically, the harmonic series).

(D) The sum of terms of an AP.

This phrase explicitly describes a series, as it refers to the sum of terms of an arithmetic progression.

Conclusion:

Option (B) $1, 4, 9, 16, ...$ is an ordered list of numbers, which is the definition of a sequence. Options (A), (C), and (D) all represent or explicitly describe a series (a sum of terms).

Therefore, option (B) is a sequence but is not necessarily presented as a series.

The correct option is (B).

Given:

The sales increase in an Arithmetic Progression (AP).

Sales in the first year ($a_1$) = $\textsf{₹} 50$ lakhs.

Sales in the $5^{th}$ year ($a_5$) = $\textsf{₹} 110$ lakhs.

To Find:

The common difference ($d$) of the AP.

Solution:

The sales figures for the years form an AP.

Let the first term be $a_1$ and the common difference be $d$.

We are given the first term, $a_1 = 50$.

The formula for the $n^{th}$ term of an AP is $a_n = a_1 + (n-1)d$.

We are given the $5^{th}$ term, $a_5 = 110$.

Using the formula for $n=5$:

$a_5 = a_1 + (5-1)d$

$a_5 = a_1 + 4d$

Substitute the given values of $a_1$ and $a_5$ into this equation:

Subtract 50 from both sides of equation (i):

$110 - 50 = 4d$

$60 = 4d$

Divide both sides by 4:

$d = \frac{60}{4}$

$d = 15$

The common difference of the AP is 15.

Since the sales are in $\textsf{₹}$ lakhs, the common difference is $\textsf{₹} 15$ lakhs.

The correct option is (C) $\textsf{₹} 15$ lakhs.

Given:

The sales increase in an Arithmetic Progression (AP).

Sales in the first year ($a_1$) = $\textsf{₹} 50$ lakhs.

From the previous question, the common difference ($d$) = $\textsf{₹} 15$ lakhs.

To Find:

The sales in the $3^{rd}$ year ($a_3$).

Solution:

The sales figures for the years form an AP with first term $a_1 = 50$ and common difference $d = 15$.

The formula for the $n^{th}$ term of an AP is $a_n = a_1 + (n-1)d$.

We need to find the sales in the $3^{rd}$ year, so we set $n = 3$.

Using the formula for $n=3$:

$a_3 = a_1 + (3-1)d$

$a_3 = a_1 + 2d$

Substitute the known values of $a_1$ and $d$:

$a_3 = 50 + 2 \times 15$

$a_3 = 50 + 30$

$a_3 = 80$

The sales in the $3^{rd}$ year were $\textsf{₹} 80$ lakhs.

The correct option is (B) $\textsf{₹} 80$ lakhs.

Given:

The sales increase in an Arithmetic Progression (AP) for the first 5 years ($n=5$).

Sales in the first year ($a_1$) = $\textsf{₹} 50$ lakhs.

Sales in the $5^{th}$ year ($a_5$) = $\textsf{₹} 110$ lakhs.

To Find:

The total sales over the first 5 years ($S_5$).

Solution:

The sales figures for the years form an AP.

We have the first term $a_1 = 50$, the last term for the sum ($a_5 = 110$), and the number of terms $n = 5$.

The formula for the sum of the first $n$ terms of an AP is given by:

$S_n = \frac{n}{2}[a_1 + a_n]$

We need to find the sum of the first 5 terms, so we set $n = 5$ and $a_n = a_5$.

Substituting the values of $a_1$, $a_5$, and $n$ into the formula, we get:

$S_5 = \frac{5}{2}[50 + 110]$

$S_5 = \frac{5}{2}[160]$

$S_5 = 5 \times \frac{160}{2}$

$S_5 = 5 \times 80$

$S_5 = 400$

The total sales over the first 5 years were $\textsf{₹} 400$ lakhs.

The correct option is (A) $\textsf{₹} 400$ lakhs.

Note: Option (B) is a duplicate of option (A).

Understanding Special Series Formulas:

Special series formulas typically refer to the standard closed-form expressions for the sum of the first $n$ terms of certain fundamental series, especially sums of powers of natural numbers.

Analyzing the Options:

(A) $\sum\limits_{k=1}^n k = 1 + 2 + 3 + ... + n$

The formula $\frac{n(n+1)}{2}$ is the well-known formula for the sum of the first $n$ natural numbers. This is a standard special series formula.

(B) $\sum\limits_{k=1}^n k^2 = 1^2 + 2^2 + 3^2 + ... + n^2$

The formula $\frac{n(n+1)(2n+1)}{6}$ is the standard formula for the sum of the squares of the first $n$ natural numbers. This is a standard special series formula.

(C) $\sum\limits_{k=1}^n k^3 = 1^3 + 2^3 + 3^3 + ... + n^3$

The formula $(\frac{n(n+1)}{2})^2$ is the standard formula for the sum of the cubes of the first $n$ natural numbers. This is a standard special series formula.

(D) $\sum\limits_{k=1}^n \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$

This represents the sum of the first $n$ terms of the Harmonic Series. While it is a significant series, it does not have a simple closed-form formula expressed as a rational function or polynomial of $n$, unlike the sums of powers of $k$. The sum is denoted by $H_n$ (the $n$-th harmonic number) and grows approximately as $\ln(n) + \gamma$, where $\gamma$ is the Euler-Mascheroni constant. It is not typically included in the set of "special series formulas" in the same context as the sums of powers of natural numbers due to the lack of a simple closed-form expression.

Conclusion:

Options (A), (B), and (C) provide standard closed-form formulas for sums of powers of natural numbers, which are commonly referred to as special series formulas. Option (D) represents the sum of the harmonic series, which does not have such a simple closed-form formula.

The correct option is (D).

Given:

A sequence whose terms are the reciprocals of the terms of an Arithmetic Progression (AP).

To Find:

The type of progression the given sequence is.

Solution:

Let the Arithmetic Progression be $a_1, a_2, a_3, ..., a_n, ...$

By the definition of an AP, the reciprocals of the terms are $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, ..., \frac{1}{a_n}, ...$

A sequence is defined to be in Harmonic Progression (HP) if the reciprocals of its terms form an Arithmetic Progression (AP).

In this question, the given sequence has terms which are the reciprocals of the terms of an AP.

Therefore, the given sequence itself is a Harmonic Progression.

The correct option is (C) HP.

Given:

The series is $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ...$

To Find:

The $n^{th}$ term of the series.

Solution:

Let the terms of the series be denoted by $a_1, a_2, a_3, ...$

The first term is $a_1 = 1 \cdot 2$.

The second term is $a_2 = 2 \cdot 3$.

The third term is $a_3 = 3 \cdot 4$.

Observing the pattern, the $k^{th}$ term of the series is the product of two consecutive natural numbers, starting with $k$.

The first factor in the $k^{th}$ term is $k$.

The second factor in the $k^{th}$ term is $k+1$.

So, the $k^{th}$ term, $a_k$, is given by $k(k+1)$.

Therefore, the $n^{th}$ term, $a_n$, is given by $n(n+1)$.

Let's verify this for the first few terms:

For $n=1$: $a_1 = 1(1+1) = 1 \cdot 2 = 2$. This matches the first term of the series ($1 \cdot 2 = 2$).

For $n=2$: $a_2 = 2(2+1) = 2 \cdot 3 = 6$. This matches the second term of the series ($2 \cdot 3 = 6$).

For $n=3$: $a_3 = 3(3+1) = 3 \cdot 4 = 12$. This matches the third term of the series ($3 \cdot 4 = 12$).

The formula $a_n = n(n+1)$ correctly represents the $n^{th}$ term of the given series.

The correct option is (A) $n(n+1)$.

Given:

The terms $a, G, b$ are in Geometric Progression (GP).

$G$ is the geometric mean of $a$ and $b$.

To Find:

The relationship between $G^2$ and $ab$.

Solution:

If $a, G, b$ are in a Geometric Progression, the ratio between consecutive terms is constant. This constant ratio is called the common ratio ($r$).

The ratio of the second term to the first term is equal to the ratio of the third term to the second term:

$\frac{G}{a} = \frac{b}{G}$

To find the relationship between $G^2$ and $ab$, we can cross-multiply this equation:

$G \times G = a \times b$

$G^2 = ab$

This shows that the square of the geometric mean of two numbers is equal to the product of the two numbers.

The correct option is (C) $G^2 = ab$.

Given:

The formula for the sum of the first $n$ terms of an AP: $S_n = \frac{n}{2}(2a + (n-1)d)$.

$l$ is the last term of the AP.

To Find:

An alternative way to write the formula for $S_n$ using the last term $l$.

Solution:

The formula for the $n^{th}$ term (or the last term, $l$) of an AP is:

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

The given formula for the sum of the first $n$ terms is:

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

We can rewrite the term $2a$ as $a + a$:

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

Now, we can group the terms $a + (n-1)d$ together:

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

Since $l = a + (n-1)d$, we can substitute $l$ into the expression:

$S_n = \frac{n}{2}(a + l)$

This matches the form given in option (A).

Now let's look at option (B): $S_n = n(a + l)/2$. This is simply a rearrangement of the formula in option (A), where the $\frac{1}{2}$ is written as division by 2.

$\frac{n}{2}(a + l) = \frac{n(a + l)}{2} = n(a + l)/2$

So, both option (A) and option (B) represent the same formula, just written slightly differently.

Option (D) $S_n = \frac{n}{2}(a + (n-1)d)$ is not the standard formula using $l$, it is a mix of $a$ and $d$ notation within the structure of the $a+l$ formula.

Since both (A) and (B) are valid alternative ways to write the sum formula using the last term $l$, the correct option is (C).

The correct option is (C) Both (A) and (B).

Given:

The $n^{th}$ term of a sequence is $a_n = 2n - 1$.

To Find:

The sum of the first $n$ terms, $S_n = \sum\limits_{k=1}^n a_k$.

Solution:

We are given the formula for the $n^{th}$ term of the sequence as $a_n = 2n - 1$.

Let's find the first few terms of the sequence:

For $n=1$: $a_1 = 2(1) - 1 = 2 - 1 = 1$

For $n=2$: $a_2 = 2(2) - 1 = 4 - 1 = 3$

For $n=3$: $a_3 = 2(3) - 1 = 6 - 1 = 5$

For $n=4$: $a_4 = 2(4) - 1 = 8 - 1 = 7$

The sequence is 1, 3, 5, 7, ...

Let's check if this sequence is an Arithmetic Progression (AP) by finding the difference between consecutive terms:

$a_2 - a_1 = 3 - 1 = 2$

$a_3 - a_2 = 5 - 3 = 2$

$a_4 - a_3 = 7 - 5 = 2$

Since the difference between consecutive terms is constant, the sequence is an AP with the first term $a = 1$ and the common difference $d = 2$.

The sum of the first $n$ terms of an AP is given by the formula:

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

Substitute the values $a = 1$ and $d = 2$ into the formula:

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

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

$S_n = \frac{n}{2}[2n]$

$S_n = n \times \frac{2n}{2}$

$S_n = n \times n$

$S_n = n^2$

The sum of the first $n$ terms of the sequence is $n^2$.

This sequence is the sequence of odd numbers, and the sum of the first $n$ odd numbers is a known result, which is $n^2$.

The correct option is (A) $n^2$.

Given:

The $n^{th}$ term of a series is $a_n$.

To Find:

The notation used for the sum of the series up to $n$ terms.

Solution:

In the context of sequences and series:

$a_n$ represents the $n^{th}$ term of the sequence/series.

$S_n$ is the standard notation commonly used to represent the sum of the first $n$ terms of a series.

The sum of the first $n$ terms of a series can be written as $a_1 + a_2 + ... + a_n$. Using summation notation, this sum is represented as $\sum\limits_{k=1}^n a_k$, where $k$ is the index of summation, starting from 1 up to $n$.

Therefore, both $S_n$ and $\sum\limits_{k=1}^n a_k$ are standard notations for the sum of the first $n$ terms of a series.

Both option (B) and option (C) are correct ways to denote the sum of the first $n$ terms of the series.

The correct option is (D) Both (B) and (C).

Understanding the Sum of an Infinite GP:

The sum of an infinite Geometric Progression (GP) with first term $a$ and common ratio $r$ is given by the formula $S_\infty = \frac{a}{1-r}$.

For this sum to be a finite, well-defined value (i.e., for the sum to exist or converge), certain conditions must be met.

Analyzing the requirements for the sum of an infinite GP to exist:

Requirement related to the common ratio $|r|$:

The formula $S_n = \frac{a(1 - r^n)}{1 - r}$ is used for the sum of the first $n$ terms. For the infinite sum to exist, the limit of $S_n$ as $n \to \infty$ must be finite.

$S_\infty = \lim\limits_{n \to \infty} \frac{a(1 - r^n)}{1 - r}$

The behavior of $r^n$ as $n \to \infty$ depends on the value of $r$:

• If $|r| < 1$, then $\lim\limits_{n \to \infty} r^n = 0$. In this case, $S_\infty = \frac{a(1 - 0)}{1 - r} = \frac{a}{1 - r}$, which is a finite value (provided $a \neq 0$).
• If $|r| > 1$, then $\lim\limits_{n \to \infty} |r^n| = \infty$. The sum diverges.
• If $r = 1$, the sequence is $a, a, a, ...$. The sum $S_n = na$, and $\lim\limits_{n \to \infty} na$ is $\infty$ (if $a \neq 0$) or 0 (if $a = 0$). For $a \neq 0$, the sum diverges.
• If $r = -1$, the sequence is $a, -a, a, -a, ...$. The sum oscillates between $a$ and 0 (if $n$ is odd) or $a$ and $a-a=0$ (if $n$ is even). The limit does not exist.

Thus, the sum of an infinite GP exists if and only if $|r| < 1$. So, option (C) is a requirement for the sum to exist.

Requirement related to the first term $a$:

If $a = 0$, the GP is $0, 0, 0, ...$. The sum of this GP is always 0, which is a finite value. So, the sum exists even if $a = 0$. However, if the question implies a non-trivial GP, $a \neq 0$ is often assumed, but it is not strictly necessary for the sum to exist (it just results in a sum of 0). If the intention is for a non-zero sum to exist, then $a$ must be non-zero. Let's re-examine the wording: "the sum of an infinite GP to exist". A sum of zero is still an existing sum. So, $a=0$ does not prevent the sum from existing.

Let's consider the standard definition of a GP: A sequence is a GP if the ratio of any term to its preceding term is a non-zero constant. This implies that $a$ cannot be zero if we want a GP with a common ratio $r$. However, some definitions allow $a=0$. If $a=0$ and $r$ is any value, the sequence is $0, 0, 0, ...$ and the sum is 0. If $a \neq 0$ and $r=0$, the sequence is $a, 0, 0, 0, ...$, the sum is $a$, which exists. But $r=0$ means $|r|<1$.

Let's assume the question is about the convergence of a standard GP with a non-zero first term for the sum formula $\frac{a}{1-r}$ to be meaningful in the non-zero case. If $a=0$, the sum is 0 regardless of $r$. So, $a \neq 0$ is required for a non-zero sum to exist.

However, let's look at option (D) first, which is more clearly NOT a requirement.

Requirement related to the common ratio $r$ being non-zero:

If $r = 0$, the GP is $a, 0, 0, 0, ...$ (assuming $a \neq 0$). The sum of the first $n$ terms is $S_n = a + 0 + ... + 0 = a$ for $n \geq 1$. The limit as $n \to \infty$ is $\lim\limits_{n \to \infty} a = a$. The sum exists and is equal to $a$. Since $r=0$, we have $|r|=0 < 1$. So, $r$ can be zero for the sum to exist. Thus, option (B) "The common ratio $r$ is non-zero" is NOT a requirement for the sum to exist.

Requirement related to the number of terms:

The question specifically asks about the sum of an infinite GP. This means the number of terms is infinite, not finite. Therefore, option (D) "The number of terms must be finite" is exactly the opposite of what is required for an infinite GP sum. For the sum of an *infinite* GP to exist, the number of terms is *infinite*, and we are asking if the *sum* of that infinite sequence converges to a finite value.

So, "The number of terms must be finite" is absolutely NOT a requirement for the sum of an *infinite* GP to exist. In fact, if the number of terms is finite, the sum always exists (it's just a finite sum).

Comparing the options with the criteria for an infinite sum to exist ($|r|<1$, and possibly $a \neq 0$ for a non-zero sum):

(A) The first term $a$ is non-zero. (If $a=0$, the sum is 0 and exists. So this is not a strict requirement for the sum to exist, but it might be assumed for a "proper" GP.)

(B) The common ratio $r$ is non-zero. (If $r=0$ and $a \neq 0$, the sum is $a$ and exists. So this is not a requirement.)

(C) The common ratio $r$ must satisfy $|r| < 1$. (This is a fundamental requirement for the sum to converge, unless $a=0$).

(D) The number of terms must be finite. (This is the opposite of the scenario we are considering - an infinite GP).

The question asks which is NOT a requirement for the sum of an infinite GP to exist.

Option (D) is clearly incorrect because we are dealing with an *infinite* GP.

Option (B) is also incorrect because the sum exists when $r=0$ (provided $a \neq 0$).

Option (A) is also incorrect because if $a=0$, the sum exists (it's 0), regardless of $r$.

However, among the given options, the most definitively incorrect statement about the requirements for the sum of an infinite GP to exist is (D). The concept of an "infinite GP" inherently means there are an infinite number of terms.

Options (A) and (B) are conditions that, if not met, might lead to a sum of zero (if $a=0$) or a GP that is trivial (if $r=0$), but the sum *does* exist in these cases.

Revisiting the common phrasing: "The sum of an infinite GP $a, ar, ar^2, ...$ converges (exists) if and only if $|r|<1$ or $a=0$."

• If $a \neq 0$, the condition is $|r|<1$. So, if $a \neq 0$, $r$ must be non-zero (unless $a=0$ is allowed, which makes the whole sequence zero).
• If $a=0$, the sum is 0 regardless of $r$.

Let's consider the options again:

(A) The first term $a$ is non-zero. (Not a requirement for the sum to exist, as sum exists if $a=0$).

(B) The common ratio $r$ is non-zero. (Not a requirement for the sum to exist if $a \neq 0$, as sum exists when $r=0$).

(C) The common ratio $r$ must satisfy $|r| < 1$. (This is a requirement if $a \neq 0$. If $a=0$, the sum is 0 regardless of $r$).

(D) The number of terms must be finite. (This is directly opposite to the problem statement being about an *infinite* GP).

The question asks which is NOT a requirement. Both (A) and (B) are not strict requirements for the sum to exist (as the sum exists and is 0 or $a$ respectively under certain conditions). However, option (D) makes no sense in the context of finding the sum of an *infinite* GP. The problem is specifically about the case where the number of terms is infinite. Therefore, requiring a finite number of terms is fundamentally incorrect for this specific problem.

Among the given choices, option (D) is the most accurate answer to "NOT a requirement for the sum of an infinite GP to exist" because the number of terms *must* be infinite for it to be an infinite GP.

The correct option is (D).

Given:

Two positive numbers, $a$ and $b$.

To Complete:

The inequality $\text{AM} \geq$ ________.

Solution:

For two positive numbers $a$ and $b$, the Arithmetic Mean (AM), Geometric Mean (GM), and Harmonic Mean (HM) are defined as:

Arithmetic Mean (AM) = $\frac{a+b}{2}$

Geometric Mean (GM) = $\sqrt{ab}$

Harmonic Mean (HM) = $\frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2ab}{a+b}$

A fundamental inequality relating these means for positive numbers states that:

$\text{AM} \geq \text{GM} \geq \text{HM}$

Equality holds if and only if $a=b$.

From this inequality, it is clear that the Arithmetic Mean (AM) is always greater than or equal to the Geometric Mean (GM).

Also, the Arithmetic Mean (AM) is always greater than or equal to the Harmonic Mean (HM).

The given statement is $\text{AM} \geq$ ________.

Based on the inequality $\text{AM} \geq \text{GM} \geq \text{HM}$, the blank can be correctly filled such that AM is greater than or equal to the quantity. Both GM and HM satisfy this condition.

Let's examine the options:

(A) GM: $\text{AM} \geq \text{GM}$ is true.

(B) HM: $\text{AM} \geq \text{HM}$ is true.

(C) Both GM and HM: This option encompasses the fact that AM is greater than or equal to both GM and HM individually.

(D) Their product $ab$: $\text{AM} \geq ab$ is not a generally true inequality for all positive numbers $a$ and $b$. For instance, if $a=0.5$ and $b=0.5$, AM = 0.5 and $ab = 0.25$, so AM > $ab$. But if $a=100$ and $b=100$, AM = 100 and $ab = 10000$, so AM < $ab$.

The most comprehensive and correct completion among the given options is that AM is greater than or equal to both GM and HM.

The correct option is (C) Both GM and HM.

Given:

Initial deposit (Principal) = $\textsf{₹} 1000$.

Interest rate = 6% compounded annually.

To Find:

The type of sequence formed by the amounts at the end of each year.

Solution:

Let the initial principal be $P_0 = \textsf{₹} 1000$.

The annual interest rate is $r = 6\% = 0.06$.

The amount ($A_n$) at the end of $n$ years with compound interest is given by the formula:

$A_n = P_0 (1 + r)^n$

The sequence of amounts at the end of each year starts from year 1. Let the amount at the end of year $n$ be $a_n$.

Amount at the end of year 1 ($a_1$) = $P_0 (1 + r)^1 = 1000 (1 + 0.06)^1 = 1000 \times 1.06 = 1060$.

Amount at the end of year 2 ($a_2$) = $P_0 (1 + r)^2 = 1000 (1.06)^2 = 1000 \times 1.1236 = 1123.60$.

Amount at the end of year 3 ($a_3$) = $P_0 (1 + r)^3 = 1000 (1.06)^3 = 1000 \times 1.191016 = 1191.016$.

The sequence of amounts is 1060, 1123.60, 1191.016, ...

Let's check if this is an Arithmetic Progression (AP) by looking at the difference between consecutive terms:

$a_2 - a_1 = 1123.60 - 1060 = 63.60$.

$a_3 - a_2 = 1191.016 - 1123.60 = 67.416$.

The difference is not constant, so it is not an AP.

Let's check if this is a Geometric Progression (GP) by looking at the ratio of consecutive terms:

$\frac{a_2}{a_1} = \frac{1123.60}{1060} = 1.06$.

$\frac{a_3}{a_2} = \frac{1191.016}{1123.60} = 1.06$.

The ratio of consecutive terms is constant, which is $1.06$.

The sequence $a_n = 1000 (1.06)^n$ for $n=1, 2, 3, ...$ is a Geometric Progression with the first term $a = 1000 \times 1.06$ and the common ratio $r = 1.06$.

The sequence formed by the amounts at the end of each year is a Geometric Progression.

The correct option is (B) GP.

Given:

The sequence of amounts at the end of each year for a deposit of $\textsf{₹} 1000$ at 6% interest compounded annually.

This sequence forms a Geometric Progression (GP) as determined in the previous question.

Initial Principal ($P_0$) = $\textsf{₹} 1000$.

Annual interest rate ($r_{rate}$) = 6% = 0.06.

To Find:

The common ratio ($r$) of this GP.

Solution:

The amount ($A_n$) at the end of $n$ years with compound interest is given by the formula:

$A_n = P_0 (1 + r_{rate})^n$

Let the amount at the end of year $n$ be $a_n$. The sequence is formed by $a_1, a_2, a_3, ...$

$a_n = 1000 (1 + 0.06)^n = 1000 (1.06)^n$

The common ratio ($r$) of a GP is the ratio of any term to its preceding term ($r = \frac{a_n}{a_{n-1}}$).

Using the formula for $a_n$:

$r = \frac{1000 (1.06)^n}{1000 (1.06)^{n-1}}$

$r = (1.06)^{n - (n-1)}$

$r = (1.06)^{1}$

$r = 1.06$

Alternatively, we can calculate the ratio using the first two terms (from the previous question):

$a_1 = 1060$

$a_2 = 1123.60$

$r = \frac{a_2}{a_1} = \frac{1123.60}{1060} = 1.06$

The common ratio of the sequence is 1.06.

The correct option is (B) 1.06.

Given:

The first term of a Geometric Progression (GP) is $a$.

The $n^{th}$ term of the GP is $l$.

To Find:

The product of the first $n$ terms of the GP.

Solution:

Let the first term of the GP be $a_1 = a$.

Let the common ratio of the GP be $r$.

The terms of the GP are $a_1, a_2, a_3, ..., a_n$, which are $a, ar, ar^2, ..., ar^{n-1}$.

The $n^{th}$ term is given as $l$. So, $l = ar^{n-1}$.

Let $P_n$ be the product of the first $n$ terms of the GP.

$P_n = a_1 \times a_2 \times a_3 \times ... \times a_n$

$P_n = a \times (ar) \times (ar^2) \times ... \times (ar^{n-1})$

We can group the terms with $a$ and the terms with $r$:

$P_n = (a \times a \times ... \times a) \times (r^0 \times r^1 \times r^2 \times ... \times r^{n-1})$

There are $n$ terms, so there are $n$ factors of $a$. Thus, the product of the $a$'s is $a^n$.

The sum of the exponents of $r$ is the sum of an arithmetic series: $0 + 1 + 2 + ... + (n-1)$. The sum of the first $k$ non-negative integers is $\frac{k(k-1)}{2}$. Here, the last term is $n-1$, so the sum of exponents is $\frac{(n-1)((n-1)+1)}{2} = \frac{(n-1)n}{2} = \frac{n(n-1)}{2}$.

So, the product of the $r$'s is $r^{\frac{n(n-1)}{2}}$.

Therefore, the product of the first $n$ terms is:

$P_n = a^n \cdot r^{\frac{n(n-1)}{2}}$

We are given that $l = ar^{n-1}$. We can write $r^{n-1} = \frac{l}{a}$.

Let's rewrite the exponent of $r$ in $P_n$:

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

So, $P_n = a^n \cdot r^{\frac{n}{2}(n-1)}$

$P_n = a^n \cdot (r^{n-1})^{\frac{n}{2}}$

Now substitute $r^{n-1} = \frac{l}{a}$:

$P_n = a^n \cdot \left(\frac{l}{a}\right)^{\frac{n}{2}}$

$P_n = a^n \cdot \frac{l^{n/2}}{a^{n/2}}$

$P_n = a^{n - n/2} \cdot l^{n/2}$

$P_n = a^{n/2} \cdot l^{n/2}$

$P_n = (al)^{n/2}$

This matches option (A).

Let's examine option (B): $\sqrt{(al)^n}$.

$\sqrt{(al)^n} = ((al)^n)^{1/2} = (al)^{n \times 1/2} = (al)^{n/2}$

So, option (B) is equivalent to option (A).

Therefore, both (A) and (B) are correct formulas for the product of the first $n$ terms.

The correct option is (D) Both (A) and (B).

Given:

The series is the sum of the squares of the first $n$ natural numbers: $1^2 + 2^2 + 3^2 + ... + n^2$.

To Find:

The formula for the sum of the first $n$ terms of this series.

Solution:

The series is the sum of the squares of the first $n$ natural numbers. This is a standard special series.

The sum of the first $n$ natural numbers is $\sum\limits_{k=1}^n k = 1 + 2 + ... + n = \frac{n(n+1)}{2}$. This is option (A).

The sum of the cubes of the first $n$ natural numbers is $\sum\limits_{k=1}^n k^3 = 1^3 + 2^3 + ... + n^3 = (\frac{n(n+1)}{2})^2$. This is option (B).

The sum of the squares of the first $n$ natural numbers is $\sum\limits_{k=1}^n k^2 = 1^2 + 2^2 + ... + n^2$.

The formula for the sum of the squares of the first $n$ natural numbers is:

$\sum\limits_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$

Comparing this with the given options, we see that it matches option (C).

The correct option is (C) $\frac{n(n+1)(2n+1)}{6}$.

Given:

The terms $a, b, c$ are in Geometric Progression (GP).

To Find:

Which of the given relationships between $a, b, c$ is true.

Solution:

If $a, b, c$ are in a Geometric Progression (GP), the ratio of any term to its preceding term is constant. This constant ratio is called the common ratio ($r$).

According to the definition of a GP, we have:

$\frac{\text{second term}}{\text{first term}} = \text{common ratio}$

$\frac{b}{a} = r$

and

$\frac{\text{third term}}{\text{second term}} = \text{common ratio}$

$\frac{c}{b} = r$

Since both ratios are equal to the common ratio $r$, they must be equal to each other:

$\frac{b}{a} = \frac{c}{b}$

To find the relationship between $a, b,$ and $c$, we can cross-multiply this equation:

$b \times b = a \times c$

$b^2 = ac$

This relationship $b^2 = ac$ is a fundamental property of terms in a Geometric Progression where $b$ is the geometric mean of $a$ and $c$.

Comparing the derived relationship $b^2 = ac$ with the given options:

(A) $a^2 = bc$ - This is incorrect.

(B) $b^2 = ac$ - This is correct.

(C) $c^2 = ab$ - This is incorrect.

(D) $b = \frac{a+c}{2}$ - This is the relationship if $a, b, c$ are in an Arithmetic Progression (AP), not a GP.

The correct option is (B) $b^2 = ac$.

Given:

The series is $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...$

To Find:

The sum of the first $n$ terms of the series.

Solution:

The given series is $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...$

Let's identify the type of progression.

The terms are $a_1 = 1$, $a_2 = \frac{1}{2}$, $a_3 = \frac{1}{4}$, $a_4 = \frac{1}{8}$, and so on.

Check the difference between consecutive terms:

$a_2 - a_1 = \frac{1}{2} - 1 = -\frac{1}{2}$

$a_3 - a_2 = \frac{1}{4} - \frac{1}{2} = \frac{1-2}{4} = -\frac{1}{4}$

The difference is not constant, so it is not an AP.

Check the ratio of consecutive terms:

$\frac{a_2}{a_1} = \frac{1/2}{1} = \frac{1}{2}$

$\frac{a_3}{a_2} = \frac{1/4}{1/2} = \frac{1}{4} \times \frac{2}{1} = \frac{2}{4} = \frac{1}{2}$

$\frac{a_4}{a_3} = \frac{1/8}{1/4} = \frac{1}{8} \times \frac{4}{1} = \frac{4}{8} = \frac{1}{2}$

The ratio of consecutive terms is constant, so the series is a Geometric Progression (GP).

The first term is $a = 1$.

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

The formula for the sum of the first $n$ terms of a GP is given by $S_n = \frac{a(1 - r^n)}{1 - r}$ (since $r \neq 1$).

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

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

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

Dividing by $\frac{1}{2}$ is the same as multiplying by 2:

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

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

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

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

Using the property of exponents $\frac{2}{2^n} = \frac{2^1}{2^n} = 2^{1-n} = \frac{1}{2^{n-1}} = \left(\frac{1}{2}\right)^{n-1}$.

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

Comparing the result with the given options:

(A) $2 - (\frac{1}{2})^{n-1}$ - Matches our derived formula.

(B) $2 - (\frac{1}{2})^n$ - Incorrect.

(C) $1 - (\frac{1}{2})^n$ - Incorrect.

(D) $1 - (\frac{1}{2})^{n-1}$ - Incorrect.

The correct option is (A) $2 - (\frac{1}{2})^{n-1}$.

Given:

The sum of the first $n$ terms of an Arithmetic Progression (AP) is given by the formula $S_n = 3n^2 + 5n$.

To Find:

The common difference ($d$) of the AP.

Solution:

The $n^{th}$ term of an AP ($a_n$) can be found using the sum of the first $n$ terms ($S_n$) and the sum of the first $(n-1)$ terms ($S_{n-1}$) with the formula:

$a_n = S_n - S_{n-1}$

We are given $S_n = 3n^2 + 5n$.

To find $S_{n-1}$, we replace $n$ with $(n-1)$ in the formula for $S_n$:

$S_{n-1} = 3(n-1)^2 + 5(n-1)$

$S_{n-1} = 3(n^2 - 2n + 1) + 5n - 5$

$S_{n-1} = 3n^2 - 6n + 3 + 5n - 5$

$S_{n-1} = 3n^2 - n - 2$

Now, we can find $a_n$:

$a_n = S_n - S_{n-1}$

$a_n = (3n^2 + 5n) - (3n^2 - n - 2)$

$a_n = 3n^2 + 5n - 3n^2 + n + 2$

$a_n = 6n + 2$

This is the formula for the $n^{th}$ term of the AP.

The general formula for the $n^{th}$ term of an AP is $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference.

We can rewrite the derived formula for $a_n$ to match the general form:

$a_n = 6n + 2$

$a_n = 6n - 6 + 8$

$a_n = 8 + 6(n-1)$

Comparing $a_n = 8 + 6(n-1)$ with $a_n = a_1 + (n-1)d$, we can identify the first term $a_1$ and the common difference $d$.

We have $a_1 = 8$ and $d = 6$.

Alternate Method:

We can find the first few terms of the AP using the sum formula $S_n = 3n^2 + 5n$.

The sum of the first 1 term ($S_1$) is the first term ($a_1$).

$S_1 = 3(1)^2 + 5(1) = 3(1) + 5 = 3 + 5 = 8$.

So, $a_1 = 8$.

The sum of the first 2 terms ($S_2$) is the sum of the first and second terms ($a_1 + a_2$).

$S_2 = 3(2)^2 + 5(2) = 3(4) + 10 = 12 + 10 = 22$.

So, $S_2 = a_1 + a_2 = 22$.

Substitute the value of $a_1$:

$8 + a_2 = 22$

$a_2 = 22 - 8$

$a_2 = 14$

The common difference ($d$) of an AP is the difference between any term and its preceding term.

$d = a_2 - a_1$

$d = 14 - 8$

$d = 6$

Both methods yield the same common difference.

The common difference of the AP is 6.

The correct option is (C) 6.

Given:

$S_n$ is the sum of the first $n$ terms of an AP.

$S_{n-1}$ is the sum of the first $(n-1)$ terms of the same AP.

$a_n$ is the $n^{th}$ term of the AP.

To Find:

The formula for the $n^{th}$ term ($a_n$) in terms of $S_n$ and $S_{n-1}$.

Solution:

The sum of the first $n$ terms of a series (including an AP) is the sum of its first $(n-1)$ terms plus the $n^{th}$ term.

Mathematically, this can be written as:

$S_n = a_1 + a_2 + ... + a_{n-1} + a_n$

The sum of the first $(n-1)$ terms is:

$S_{n-1} = a_1 + a_2 + ... + a_{n-1}$

To find the $n^{th}$ term $a_n$, we can subtract the sum of the first $(n-1)$ terms from the sum of the first $n$ terms:

$S_n - S_{n-1} = (a_1 + a_2 + ... + a_{n-1} + a_n) - (a_1 + a_2 + ... + a_{n-1})$

$S_n - S_{n-1} = a_n$

Thus, the $n^{th}$ term of an AP (or any sequence where the sum of terms is defined) is given by the difference between the sum of the first $n$ terms and the sum of the first $(n-1)$ terms.

$a_n = S_n - S_{n-1}$

This formula holds true for $n > 1$. For $n=1$, the first term $a_1$ is simply $S_1$. Using the formula, $a_1 = S_1 - S_0$. Assuming $S_0 = 0$ (the sum of zero terms is zero), the formula is also consistent for $n=1$.

Comparing the derived formula with the given options:

(A) $S_n - S_{n-1}$ - Matches our result.

(B) $S_n + S_{n-1}$ - Incorrect.

(C) $S_n - S_n = 0$ - Incorrect (unless the $n^{th}$ term is 0).

(D) $a + (n-1)d$ - This is the formula for the $n^{th}$ term in terms of the first term and common difference, not in terms of $S_n$ and $S_{n-1}$.

The correct option is (A) $S_n - S_{n-1}$.

Assertion (A): The geometric mean of two positive numbers is always less than or equal to their arithmetic mean.

Reason (R): For positive numbers $a, b$, $\sqrt{ab} \leq \frac{a+b}{2}$. Equality holds when $a=b$.

Analysis of Assertion (A):

For any two positive numbers $a$ and $b$, the Arithmetic Mean (AM) is $\frac{a+b}{2}$ and the Geometric Mean (GM) is $\sqrt{ab}$. The fundamental inequality between AM and GM states that the AM is always greater than or equal to the GM.

$\frac{a+b}{2} \geq \sqrt{ab}$

This is equivalent to stating that the Geometric Mean is always less than or equal to the Arithmetic Mean ($\sqrt{ab} \leq \frac{a+b}{2}$).

Thus, Assertion (A) is true.

Analysis of Reason (R):

Reason (R) directly provides the mathematical inequality $\sqrt{ab} \leq \frac{a+b}{2}$ for positive numbers $a$ and $b$, and correctly states that equality holds when $a=b$. This inequality is the precise mathematical formulation of the relationship between the geometric mean and the arithmetic mean of two positive numbers.

Thus, Reason (R) is true.

Relationship between Assertion (A) and Reason (R):

Assertion (A) makes a statement about the relationship between the geometric mean and the arithmetic mean. Reason (R) provides the exact mathematical inequality that represents this relationship, along with the condition for equality. Reason (R) is a precise mathematical statement that directly supports and explains the assertion made in (A).

Therefore, Reason (R) is the correct explanation for Assertion (A).

Conclusion:

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Given:

An Arithmetic Progression (AP).

The $p^{th}$ term of the AP is $a$.

The $q^{th}$ term of the AP is $b$.

The $r^{th}$ term of the AP is $c$.

To Find:

The value of the expression $a(q-r) + b(r-p) + c(p-q)$.

Solution:

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

The formula for the $n^{th}$ term of an AP is $T_n = A + (n-1)D$.

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

$a = T_p = A + (p-1)D$

$b = T_q = A + (q-1)D$

$c = T_r = A + (r-1)D$

Now, substitute these expressions for $a, b, c$ into the given expression:

Expression = $a(q-r) + b(r-p) + c(p-q)$

Expression = $[A + (p-1)D](q-r) + [A + (q-1)D](r-p) + [A + (r-1)D](p-q)$

Expand each term by distributing the factors $(q-r)$, $(r-p)$, and $(p-q)$: orthodoxy

Expression = $A(q-r) + (p-1)D(q-r) + A(r-p) + (q-1)D(r-p) + A(p-q) + (r-1)D(p-q)$

Group the terms containing $A$ and the terms containing $D$:

Expression = $[A(q-r) + A(r-p) + A(p-q)] + [(p-1)D(q-r) + (q-1)D(r-p) + (r-1)D(p-q)]

Factor out $A$ from the first group and $D$ from the second group:

Expression = $A[(q-r) + (r-p) + (p-q)] + D[(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)]

Simplify the terms inside the square brackets:

Consider the first bracket: $(q-r) + (r-p) + (p-q) = q - r + r - p + p - q = (q-q) + (-r+r) + (-p+p) = 0$

Consider the second bracket. Expand the products:

$(p-1)(q-r) = pq - pr - q + r$

$(q-1)(r-p) = qr - qp - r + p$

$(r-1)(p-q) = rp - rq - p + q$

Sum these three expanded terms:

$(pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q)

= $pq - pr - q + r + qr - pq - r + p + rp - rq - p + q$

= $(pq - pq) + (-pr + rp) + (-q + q) + (r - r) + (qr - rq) + (p - p)$

= $0 + 0 + 0 + 0 + 0 + 0 = 0$

Substitute the simplified results back into the expression:

Expression = $A[0] + D[0]$

Expression = $0 + 0$

Expression = $0$

The value of the expression $a(q-r) + b(r-p) + c(p-q)$ is 0.

The correct option is (A) 0.

Solution:

The sum of the cubes of the first $n$ natural numbers is given by the formula:

$\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$

In this problem, we need to find the sum of the series $1^3 + 2^3 + 3^3 + ... + 10^3$. Here, $n=10$.

Using the formula with $n=10$:

Sum $= \left(\frac{10(10+1)}{2}\right)^2$

Sum $= \left(\frac{10 \times 11}{2}\right)^2$

Sum $= \left(\frac{110}{2}\right)^2$

Sum $= (55)^2$

Sum $= 55 \times 55$

Sum $= 3025$

The sum of the series $1^3 + 2^3 + 3^3 + ... + 10^3$ is 3025.

The correct option is (A) 3025.

Solution:

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by $d$.

Let's examine each option:

(A) 2, 4, 6, 8

Difference between the 2nd and 1st term: $4 - 2 = 2$

Difference between the 3rd and 2nd term: $6 - 4 = 2$

Difference between the 4th and 3rd term: $8 - 6 = 2$

Since the difference between consecutive terms is constant (which is 2), this sequence is an AP.

(B) 1, 2, 4, 8

Difference between the 2nd and 1st term: $2 - 1 = 1$

Difference between the 3rd and 2nd term: $4 - 2 = 2$

Since the difference between consecutive terms is not constant ($1 \ne 2$), this sequence is not an AP. (This is a Geometric Progression with a common ratio of 2).

(C) 5, 5, 5, 5

Difference between the 2nd and 1st term: $5 - 5 = 0$

Difference between the 3rd and 2nd term: $5 - 5 = 0$

Difference between the 4th and 3rd term: $5 - 5 = 0$

Since the difference between consecutive terms is constant (which is 0), this sequence is an AP.

(D) $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}$

Difference between the 2nd and 1st term: $\frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = -\frac{1}{6}$

Difference between the 3rd and 2nd term: $\frac{1}{4} - \frac{1}{3} = \frac{3 - 4}{12} = -\frac{1}{12}$

Since the difference between consecutive terms is not constant ($-\frac{1}{6} \ne -\frac{1}{12}$), this sequence is not an AP. As mentioned in the question, the reciprocals $2, 3, 4, 5$ form an AP, so this sequence is a Harmonic Progression (HP).

Based on the analysis, options (A) and (C) are sequences that can be terms of an AP.

The correct options are (A) and (C).

Solution:

The given formula for the sum of the first $n$ terms of a Geometric Progression (GP) with first term $a$ and common ratio $r$ (where $r \neq 1$) is:

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

Let's examine the given options for equivalence.

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

Consider the given formula: $S_n = a \frac{(r^n - 1)}{r - 1}$.

We can multiply the numerator and the denominator by -1:

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

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

This matches option (A). So, option (A) is an equivalent formula.

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

Consider the given formula: $S_n = a \frac{(r^n - 1)}{r - 1}$.

We can distribute the term $a$ into the parenthesis in the numerator:

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

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

This matches option (B). So, option (B) is an equivalent formula.

(C) Both (A) and (B)

Since both option (A) and option (B) are shown to be equivalent to the given formula, option (C) is correct.

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

This formula is for the sum of the first $n$ terms of an Arithmetic Progression (AP), not a Geometric Progression (GP). Thus, it is not equivalent to the given formula for the sum of a GP.

Therefore, both formulas in options (A) and (B) are equivalent to the given formula.

The correct option is (C) Both (A) and (B).

Solution:

Let the first AP have first term $a_1$ and common difference $d_1$.

Let the second AP have first term $a_2$ and common difference $d_2$.

The sum of the first $n$ terms of an AP is given by the formula $S_n = \frac{n}{2}[2a + (n-1)d]$.

For the first AP, the sum of the first $n$ terms is $S_n^{(1)} = \frac{n}{2}[2a_1 + (n-1)d_1]$.

For the second AP, the sum of the first $n$ terms is $S_n^{(2)} = \frac{n}{2}[2a_2 + (n-1)d_2]$.

We are given that the ratio of the sum of the first $n$ terms of the two APs is $(2n+1) : (2n-1)$.

$\frac{S_n^{(1)}}{S_n^{(2)}} = \frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2}$

So, we have:

$\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{2n+1}{2n-1}$

We need to find the ratio of their $10^{th}$ terms. The $k^{th}$ term of an AP is given by $a_k = a + (k-1)d$.

The $10^{th}$ term of the first AP is $a_{10}^{(1)} = a_1 + (10-1)d_1 = a_1 + 9d_1$.

The $10^{th}$ term of the second AP is $a_{10}^{(2)} = a_2 + (10-1)d_2 = a_2 + 9d_2$.

We want to find the ratio $\frac{a_{10}^{(1)}}{a_{10}^{(2)}} = \frac{a_1 + 9d_1}{a_2 + 9d_2}$.

Let's look at the expression we derived from the sum ratio: $\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2}$.

We can factor out 2 from the numerator and denominator:

$\frac{2(a_1 + \frac{n-1}{2}d_1)}{2(a_2 + \frac{n-1}{2}d_2)} = \frac{a_1 + \frac{n-1}{2}d_1}{a_2 + \frac{n-1}{2}d_2}$

So, $\frac{a_1 + \frac{n-1}{2}d_1}{a_2 + \frac{n-1}{2}d_2} = \frac{2n+1}{2n-1}$.

We want the ratio $\frac{a_1 + 9d_1}{a_2 + 9d_2}$.

Comparing the required ratio with the expression we have, we see that if we choose a value of $n$ such that $\frac{n-1}{2} = 9$, the expression will match the required term ratio.

Let's solve for $n$:

$\frac{n-1}{2} = 9$

$n-1 = 9 \times 2$

$n-1 = 18$

$n = 18 + 1$

$n = 19$

Now, substitute $n=19$ into the given ratio of sums formula:

$\frac{S_{19}^{(1)}}{S_{19}^{(2)}} = \frac{2(19)+1}{2(19)-1}$

$\frac{S_{19}^{(1)}}{S_{19}^{(2)}} = \frac{38+1}{38-1}$

$\frac{S_{19}^{(1)}}{S_{19}^{(2)}} = \frac{39}{37}$

We know that for $n=19$:

$\frac{S_{19}^{(1)}}{S_{19}^{(2)}} = \frac{a_1 + \frac{19-1}{2}d_1}{a_2 + \frac{19-1}{2}d_2} = \frac{a_1 + 9d_1}{a_2 + 9d_2}$

Thus, the ratio of the $10^{th}$ terms is equal to the ratio of the sums when $n=19$.

$\frac{a_{10}^{(1)}}{a_{10}^{(2)}} = \frac{39}{37}$

The ratio of their $10^{th}$ terms is $39 : 37$.

The correct option is (C) 39 : 37.

Solution:

Let the first term of the Geometric Progression (GP) be $a$ and the common ratio be $r$.

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

We are given the first term, $a = 729$.

We are also given the $7^{th}$ term, $a_7 = 64$.

Using the formula for the $n^{th}$ term, the $7^{th}$ term is $a_7 = ar^{7-1} = ar^6$.

Substitute the given values into the equation:

$64 = 729 \times r^6$

Now, we need to solve for $r$. Divide both sides by 729:

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

To find $r$, we need to take the $6^{th}$ root of both sides.

$r = \pm \left(\frac{64}{729}\right)^{\frac{1}{6}}$

We need to find the base numbers which, when raised to the power of 6, give 64 and 729.

$2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$

$3^6 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 9 \times 9 \times 9 = 81 \times 9 = 729$

So, $\left(\frac{64}{729}\right)^{\frac{1}{6}} = \left(\frac{2^6}{3^6}\right)^{\frac{1}{6}} = \left(\left(\frac{2}{3}\right)^6\right)^{\frac{1}{6}} = \frac{2}{3}$.

Therefore, the common ratio $r$ can be positive or negative:

$r = \pm \frac{2}{3}$

The common ratio is $\pm \frac{2}{3}$.

The correct option is (C) $\pm 2/3$.

Solution:

The sum of an infinite Geometric Progression (GP) with first term $a$ and common ratio $r$ is given by the formula $S_\infty = \frac{a}{1-r}$, provided that the absolute value of the common ratio $|r|$ is less than 1.

The condition for the sum of an infinite GP to converge to a finite value is $|r| < 1$, which means $-1 < r < 1$.

If $|r| \geq 1$ and $r \neq 1$, the terms of the GP do not approach zero, and the sum of the series diverges (does not approach a finite value). If $r = 1$, the sum of the infinite series is undefined (unless $a=0$).

The question states that the common ratio of a GP is between -1 and 1 (exclusive), which is exactly the condition $|r| < 1$.

Under this condition, the sum of the infinite GP converges to the finite value $\frac{a}{1-r}$ (assuming $a \neq 0$).

Therefore, if the common ratio of a GP is between -1 and 1 (exclusive), the sum of the infinite GP converges to a finite value.

The correct option is (C) converges to a finite value.

Solution:

The given series is $1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + ... + (2n-1)(2n+1)$.

The terms of the series are formed by the product of consecutive odd numbers. The $k^{th}$ term of the series, denoted by $a_k$, is given by the general expression $(2k-1)(2k+1)$, where $k$ goes from 1 to $n$.

We expand the $k^{th}$ term:

$a_k = (2k-1)(2k+1)$

This is a difference of squares $(A-B)(A+B) = A^2 - B^2$ where $A=2k$ and $B=1$.

$a_k = (2k)^2 - 1^2 = 4k^2 - 1$

The sum of the first $n$ terms of the series, $S_n$, is the sum of the $k^{th}$ terms from $k=1$ to $n$:

$S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (4k^2 - 1)$

Using the linearity property of summation ($\sum (c a_k \pm d b_k) = c \sum a_k \pm d \sum b_k$), we split the sum:

$S_n = \sum_{k=1}^{n} 4k^2 - \sum_{k=1}^{n} 1$

We can take the constant factor 4 outside the summation:

$S_n = 4 \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} 1$

We use the standard formulas for the sum of the first $n$ squares ($\sum_{k=1}^{n} k^2$) and the sum of $n$ ones ($\sum_{k=1}^{n} 1$):

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^{n} 1 = n$

Substitute these formulas into the expression for $S_n$:

$S_n = 4 \left(\frac{n(n+1)(2n+1)}{6}\right) - n$

Simplify the constant factor $\frac{4}{6}$ in the first term to $\frac{2}{3}$:

$S_n = \frac{4}{6} n(n+1)(2n+1) - n$

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

To express this sum as a single fraction, we find a common denominator, which is 3:

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

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

Now, we expand the product $(n+1)(2n+1)$ in the numerator:

$(n+1)(2n+1) = n(2n+1) + 1(2n+1) = 2n^2 + n + 2n + 1 = 2n^2 + 3n + 1$

Substitute this back into the numerator:

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

Distribute $2n$ into the parenthesis:

$S_n = \frac{4n^3 + 6n^2 + 2n - 3n}{3}$

Combine the like terms ($2n$ and $-3n$) in the numerator:

$S_n = \frac{4n^3 + 6n^2 - n}{3}$

Finally, factor out $n$ from the terms in the numerator:

$S_n = \frac{n (4n^2 + 6n - 1)}{3}$

Comparing this result with the given options, we see that it matches option (B).

Note that option (A) represents the sum in an intermediate step of the calculation, which is also mathematically correct. Option (D) shows a calculation step that leads to the simplified form presented in option (B), confirming its correctness.

Assuming the question asks for the simplified form of the sum as a single expression, option (B) is the intended answer.

The correct option is (B) $\frac{n(4n^2+6n-1)}{3}$.

Solution:

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

The Arithmetic Mean (AM) of $a$ and $b$ is given by $\text{AM} = \frac{a+b}{2}$.

The Geometric Mean (GM) of $a$ and $b$ is given by $\text{GM} = \sqrt{ab}$.

The Harmonic Mean (HM) of $a$ and $b$ is given by $\text{HM} = \frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2ab}{a+b}$.

For two positive numbers, there is a relationship between AM, GM, and HM given by $\text{GM}^2 = \text{AM} \times \text{HM}$.

We are given that $\text{AM} = 5$ and $\text{GM} = 4$.

We need to find the HM. Using the relationship $GM^2 = AM \times HM$:

$(4)^2 = 5 \times \text{HM}$

$16 = 5 \times \text{HM}$

To find HM, divide both sides by 5:

$\text{HM} = \frac{16}{5}$

Comparing the calculated value with the given options:

(A) $4^2 / 5 = 16/5$. This matches our result.

(B) $5 \times 4 / 2 = 20 / 2 = 10$. This is incorrect.

(C) $\sqrt{5 \times 4} = \sqrt{20}$. This is the Geometric Mean of 5 and 4, not the Harmonic Mean of the original two numbers.

(D) $5/4$. This is incorrect.

The Harmonic Mean is $\frac{16}{5}$.

The correct option is (A) $4^2 / 5 = 16/5$.

Solution:

Let the given series be denoted by $S$. The terms of the series are:

The first term is $T_1 = 1$.

The second term is $T_2 = 1 + 2$.

The third term is $T_3 = 1 + 2 + 3$.

Following this pattern, the $n^{th}$ term of the series, $T_n$, is the sum of the first $n$ natural numbers.

$T_n = 1 + 2 + 3 + ... + n$

The sum of the first $n$ natural numbers is given by the formula:

$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

Therefore, the $n^{th}$ term of the given series is:

$T_n = \frac{n(n+1)}{2}$

The $n^{th}$ term of the series is $\frac{n(n+1)}{2}$.

The correct option is (B) $n(n+1)/2$.

Solution:

For three numbers $a, b, c$ to be in an Arithmetic Progression (AP), the difference between consecutive terms must be constant.

This means that the difference between the second term ($b$) and the first term ($a$) must be equal to the difference between the third term ($c$) and the second term ($b$).

So, the condition is:

$b - a = c - b$

To rearrange this equation, we can add $b$ to both sides:

$b - a + b = c - b + b$

$2b - a = c$

Now, add $a$ to both sides:

$2b - a + a = c + a$

$2b = a + c$

This is the standard condition for three numbers to be in AP. The middle term is the arithmetic mean of the other two terms.

Let's look at the other options:

(A) $b^2 = ac$: This is the condition for $a, b, c$ to be in a Geometric Progression (GP). Here, the middle term $b$ is the geometric mean of $a$ and $c$ (for positive $a, c$, $b = \sqrt{ac}$).

(C) $\frac{b}{a} = \frac{c}{b}$: This is equivalent to $b^2 = ac$ (by cross-multiplication), which is also the condition for a GP.

(D) $\frac{1}{b} = \frac{1}{a} + \frac{1}{c}$: This is not the standard condition for any common progression type. The condition for $a, b, c$ to be in a Harmonic Progression (HP) is that their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in AP. The condition for $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ to be in AP is $\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$, which simplifies to $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$, or $b = \frac{2ac}{a+c}$. Option (D) is different from this.

Based on the definition of an AP, the correct condition is $2b = a+c$.

The correct option is (B) $2b = a+c$.

Solution:

Let the first term of the Geometric Progression (GP) be $a$ and the common ratio be $r$.

The sum of the first $n$ terms of a GP is given by the formula:

$S_n = a \frac{(r^n - 1)}{r - 1}$, for $r \neq 1$.

We are given that the sum of the first 5 terms is 242, so $n=5$ and $S_5 = 242$.

The common ratio is given as $r = 3$.

Since $r=3 \neq 1$, we can use the formula. Substitute the given values into the formula for $S_5$:

$S_5 = a \frac{(3^5 - 1)}{3 - 1}$

$242 = a \frac{(243 - 1)}{2}$

$242 = a \frac{242}{2}$

$242 = a \times 121$

Now, solve for $a$:

$a = \frac{242}{121}$

$a = 2$

The first term of the GP is 2.

The correct option is (B) 2.

Solution:

Let the first Arithmetic Progression (AP) have first term $a_1$ and common difference $d_1$.

Let the second AP have first term $a_2$ and common difference $d_2$.

The $n^{th}$ term of an AP is given by the formula $T_n = a + (n-1)d$.

For the first AP, the $n^{th}$ term is $T_n^{(1)} = a_1 + (n-1)d_1$.

For the second AP, the $n^{th}$ term is $T_n^{(2)} = a_2 + (n-1)d_2$.

We are given that the ratio of their $n^{th}$ terms is $(3n+4) : (5n-1)$.

$\frac{T_n^{(1)}}{T_n^{(2)}} = \frac{a_1 + (n-1)d_1}{a_2 + (n-1)d_2} = \frac{3n+4}{5n-1}$

We can rewrite the $n^{th}$ term formula as $T_n = a + nd - d = dn + (a-d)$. This shows that the $n^{th}$ term is a linear function of $n$.

So, $\frac{d_1 n + (a_1 - d_1)}{d_2 n + (a_2 - d_2)} = \frac{3n+4}{5n-1}$.

Since this equality holds for all values of $n$, the coefficients of the powers of $n$ in the numerator and denominator must be proportional.

Comparing the coefficients of $n$ in the numerator and denominator on both sides, we have:

$\frac{\text{Coefficient of } n \text{ in numerator 1}}{\text{Coefficient of } n \text{ in denominator 1}} = \frac{d_1}{d_2}$

$\frac{\text{Coefficient of } n \text{ in numerator 2}}{\text{Coefficient of } n \text{ in denominator 2}} = \frac{3}{5}$

Equating these ratios, we get:

$\frac{d_1}{d_2} = \frac{3}{5}$

The ratio of their common differences is $3 : 5$.

Alternatively, consider multiplying both sides of the ratio equation by the denominators:

$(a_1 + (n-1)d_1)(5n-1) = (a_2 + (n-1)d_2)(3n+4)$

$(d_1 n + a_1 - d_1)(5n - 1) = (d_2 n + a_2 - d_2)(3n + 4)$

Expanding both sides and collecting terms by powers of $n$:

$5d_1 n^2 + (5a_1 - 5d_1 - d_1) n + (d_1 - a_1) = 3d_2 n^2 + (3a_2 - 3d_2 + 4d_2) n + (4a_2 - 4d_2)$

$5d_1 n^2 + (5a_1 - 6d_1) n + (d_1 - a_1) = 3d_2 n^2 + (3a_2 + d_2) n + (4a_2 - 4d_2)$

For this polynomial equation to hold for all values of $n$, the coefficients of corresponding powers of $n$ on both sides must be equal.

Comparing the coefficients of $n^2$:

$5d_1 = 3d_2$

Dividing both sides by $5d_2$ (assuming $d_2 \neq 0$, otherwise the terms would be constant and the ratio of non-constant expressions would not hold):

$\frac{d_1}{d_2} = \frac{3}{5}$

The ratio of the common differences is $3:5$.

Looking at the options, option (A) is 3:5, and option (B) is 6:10 = 3:5. Both represent the same ratio. Option (B) explicitly states that the ratio 6:10 is equal to 3:5, making it a potentially intended answer that highlights an equivalent ratio. The simplest form of the ratio is 3:5.

The correct option is (B) 6 : 10 = 3 : 5.

Solution:

The given series is $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... + n(n+1)$.

The $k^{th}$ term of the series, denoted by $a_k$, is given by the general expression $k(k+1)$.

We can expand the $k^{th}$ term:

$a_k = k(k+1) = k^2 + k$

The sum of the first $n$ terms of the series, $S_n$, is the sum of the $k^{th}$ terms from $k=1$ to $n$:

$S_n = \sum\limits_{k=1}^{n} a_k = \sum\limits_{k=1}^{n} (k^2 + k)$

Using the linearity property of summation ($\sum (a_k \pm b_k) = \sum a_k \pm \sum b_k$), we split the sum:

$S_n = \sum\limits_{k=1}^{n} k^2 + \sum\limits_{k=1}^{n} k$

We use the standard formulas for the sum of the first $n$ squares and the sum of the first $n$ natural numbers:

$\sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum\limits_{k=1}^{n} k = \frac{n(n+1)}{2}$

Substitute these formulas into the expression for $S_n$:

$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$

To combine these terms into a single fraction, find a common denominator, which is 6:

$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1) \times 3}{2 \times 3}$

$S_n = \frac{n(n+1)(2n+1) + 3n(n+1)}{6}$

Factor out the common term $n(n+1)$ from the terms in the numerator:

$S_n = \frac{n(n+1) [ (2n+1) + 3 ]}{6}$

Simplify the expression inside the square brackets:

$S_n = \frac{n(n+1) (2n+4)}{6}$

Factor out 2 from the term $(2n+4)$:

$S_n = \frac{n(n+1) \times 2(n+2)}{6}$

Cancel the common factor of 2 in the numerator and denominator:

$S_n = \frac{\cancel{2} n(n+1)(n+2)}{\cancel{6}_{3}}$

$S_n = \frac{n(n+1)(n+2)}{3}$

The sum of the series $1 \cdot 2 + 2 \cdot 3 + ... + n(n+1)$ is $\frac{n(n+1)(n+2)}{3}$.

Comparing this result with the given options, we see that it matches option (B). Option (A) shows an intermediate step in the calculation, which is also correct but not the simplified closed form.

The correct option is (B) $\frac{n(n+1)(n+2)}{3}$.

Solution:

We are given that $a, b, c$ are in Arithmetic Progression (AP).

If three numbers are in AP, the difference between consecutive terms is constant. Let the common difference be $d$.

So, $b - a = d$ and $c - b = d$.

From these relationships, we can express the exponents in the given expression in terms of $d$:

$b-c = -(c-b) = -d$

$c-a = (c-b) + (b-a) = d + d = 2d$

$a-b = -(b-a) = -d$

The given expression is $x^{b-c} y^{c-a} z^{a-b}$. Substituting the exponents:

Expression $= x^{-d} y^{2d} z^{-d}$

We are also given that $x, y, z$ are in Geometric Progression (GP).

If three non-zero numbers are in GP, the ratio between consecutive terms is constant. Let the common ratio be $r$.

So, $\frac{y}{x} = r$ and $\frac{z}{y} = r$.

This implies $y = xr$ and $z = yr = (xr)r = xr^2$.

Another property of three terms in GP is that the square of the middle term is equal to the product of the first and third terms:

$y^2 = xz$

Now, let's rewrite the expression using the property $y^2 = xz$:

Expression $= x^{-d} y^{2d} z^{-d}$

We can group the terms with the same exponent $-d$:

Expression $= (x z)^{-d} y^{2d}$

Using the property of exponents $(A^m)^n = A^{mn}$, we can write $y^{2d}$ as $(y^2)^d$:

Expression $= (xz)^{-d} (y^2)^d$

Now, substitute $y^2 = xz$ into the expression:

Expression $= (xz)^{-d} (xz)^d$

Using the property $A^m A^n = A^{m+n}$:

Expression $= (xz)^{-d + d}$

Expression $= (xz)^0$

Any non-zero number raised to the power of 0 is 1. Assuming $x$ and $z$ are such that $xz \neq 0$:

Expression $= 1$

The value of the expression $x^{b-c} y^{c-a} z^{a-b}$ is 1.

The correct option is (B) 1.

Solution:

The given sequence is $\frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{20}, ...$.

Let's examine the denominators of the terms: 2, 6, 12, 20, ...

Let the $n^{th}$ term of the sequence be $T_n$. Then $T_n = \frac{1}{D_n}$, where $D_n$ is the $n^{th}$ denominator.

The sequence of denominators is $D_1=2, D_2=6, D_3=12, D_4=20, ...$

Let's look for a pattern in the sequence of denominators.

$D_1 = 2 = 1 \times 2$

$D_2 = 6 = 2 \times 3$

$D_3 = 12 = 3 \times 4$

$D_4 = 20 = 4 \times 5$

It appears that the $n^{th}$ denominator is the product of $n$ and $(n+1)$.

So, the formula for the $n^{th}$ denominator is $D_n = n(n+1)$.

Therefore, the $n^{th}$ term of the sequence is:

$T_n = \frac{1}{D_n} = \frac{1}{n(n+1)}$

Let's verify this formula with the first few terms:

For $n=1$, $T_1 = \frac{1}{1(1+1)} = \frac{1}{1 \times 2} = \frac{1}{2}$. (Matches the first term)

For $n=2$, $T_2 = \frac{1}{2(2+1)} = \frac{1}{2 \times 3} = \frac{1}{6}$. (Matches the second term)

For $n=3$, $T_3 = \frac{1}{3(3+1)} = \frac{1}{3 \times 4} = \frac{1}{12}$. (Matches the third term)

For $n=4$, $T_4 = \frac{1}{4(4+1)} = \frac{1}{4 \times 5} = \frac{1}{20}$. (Matches the fourth term)

The formula $\frac{1}{n(n+1)}$ correctly generates the terms of the sequence.

Comparing our result with the given options, we find that option (A) matches the derived formula for the $n^{th}$ term.

The correct option is (A) $\frac{1}{n(n+1)}$.

Solution:

A sequence of non-zero numbers is said to be in Harmonic Progression (HP) if the reciprocals of its terms are in Arithmetic Progression (AP).

We need to examine the reciprocals of the terms in each given option and check if they form an AP. An AP has a constant difference between consecutive terms.

(A) 2, 4, 6, 8

Reciprocals: $\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{8}$

Difference between 2nd and 1st term: $\frac{1}{4} - \frac{1}{2} = \frac{1 - 2}{4} = -\frac{1}{4}$

Difference between 3rd and 2nd term: $\frac{1}{6} - \frac{1}{4} = \frac{2 - 3}{12} = -\frac{1}{12}$

Since $-\frac{1}{4} \neq -\frac{1}{12}$, the reciprocals are not in AP. Thus, the sequence is not in HP.

(B) $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}$

Reciprocals: $\frac{1}{1}, \frac{1}{\frac{1}{2}}, \frac{1}{\frac{1}{3}}, \frac{1}{\frac{1}{4}}$, which simplifies to $1, 2, 3, 4$.

Difference between 2nd and 1st term: $2 - 1 = 1$

Difference between 3rd and 2nd term: $3 - 2 = 1$

Difference between 4th and 3rd term: $4 - 3 = 1$

Since the difference between consecutive terms is constant (1), the reciprocals are in AP. Thus, the sequence $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ is in HP.

(C) 3, 6, 12, 24

Reciprocals: $\frac{1}{3}, \frac{1}{6}, \frac{1}{12}, \frac{1}{24}$

Difference between 2nd and 1st term: $\frac{1}{6} - \frac{1}{3} = \frac{1 - 2}{6} = -\frac{1}{6}$

Difference between 3rd and 2nd term: $\frac{1}{12} - \frac{1}{6} = \frac{1 - 2}{12} = -\frac{1}{12}$

Since $-\frac{1}{6} \neq -\frac{1}{12}$, the reciprocals are not in AP. Thus, the sequence is not in HP. (This sequence is in GP).

(D) $1, \frac{1}{4}, \frac{1}{9}, \frac{1}{16}$

Reciprocals: $\frac{1}{1}, \frac{1}{\frac{1}{4}}, \frac{1}{\frac{1}{9}}, \frac{1}{\frac{1}{16}}$, which simplifies to $1, 4, 9, 16$.

Difference between 2nd and 1st term: $4 - 1 = 3$

Difference between 3rd and 2nd term: $9 - 4 = 5$

Since $3 \neq 5$, the reciprocals are not in AP. Thus, the sequence is not in HP.

Only the sequence in option (B) has reciprocals that form an Arithmetic Progression.

The correct option is (B) $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}$.

Solution:

The sum of the first $n$ terms of a sequence is denoted by $S_n$.

$S_n = a_1 + a_2 + a_3 + ... + a_{n-1} + a_n$

The sum of the first $(n-1)$ terms is:

$S_{n-1} = a_1 + a_2 + a_3 + ... + a_{n-1}$

Subtracting $S_{n-1}$ from $S_n$ for $n > 1$:

$S_n - S_{n-1} = (a_1 + a_2 + ... + a_{n-1} + a_n) - (a_1 + a_2 + ... + a_{n-1})$

$S_n - S_{n-1} = a_n$

This confirms the given relationship $a_n = S_n - S_{n-1}$ for $n > 1$.

We need to find the value of the first term, $a_1$.

The sum of the first 1 term of the sequence, $S_1$, is simply the first term itself.

$S_1 = a_1$

Therefore, the first term $a_1$ is equal to $S_1$.

Let's check the options:

(A) $S_1$: This matches our derivation.

(B) $S_2 - S_1$: According to the formula $a_n = S_n - S_{n-1}$, for $n=2$, $a_2 = S_2 - S_1$. This is the second term, not the first.

(C) $S_0$: $S_0$ is typically defined as the sum of zero terms, which is 0. This is not equal to $a_1$ in general.

(D) $a$: The symbol '$a$' is often used to denote the first term ($a_1$) of an AP. While $a_1$ is indeed the first term, option (A) expresses $a_1$ in terms of the sum notation $S_n$, which is consistent with the premise of the question.

In the context of relating the terms to the sums $S_n$, the most direct and correct answer for $a_1$ is $S_1$.

The correct option is (A) $S_1$.

Solution:

The sum of the first $n$ cubes is given by the formula:

$\sum\limits_{k=1}^n k^3 = 1^3 + 2^3 + 3^3 + ... + n^3 = \left(\frac{n(n+1)}{2}\right)^2$

The sum of the first $n$ natural numbers is given by the formula:

$\sum\limits_{k=1}^n k = 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$

The statement given in the question is that the sum of the series $1^3 + 2^3 + ... + n^3$ is equal to $(\sum\limits_{k=1}^n k)^2$.

Let's check if the formulas are consistent with the statement.

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

The sum $\sum\limits_{k=1}^n k$ is $\frac{n(n+1)}{2}$.

The square of the sum $\left(\sum\limits_{k=1}^n k\right)^2$ is $\left(\frac{n(n+1)}{2}\right)^2$.

Comparing the sum of the series with the square of the sum of natural numbers:

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

The equality holds true for all positive integer values of $n$. This is a well-known identity.

Therefore, the statement is true.

The correct option is (A) Yes.

Solution:

For two positive integers $a$ and $b$, the Arithmetic Mean (AM) is given by $\text{AM} = \frac{a+b}{2}$.

The Geometric Mean (GM) is given by $\text{GM} = \sqrt{ab}$.

The AM-GM inequality states that for any non-negative numbers, $\text{AM} \geq \text{GM}$. For positive numbers $a$ and $b$, this means:

$\frac{a+b}{2} \geq \sqrt{ab}$

We are asked for the condition under which equality holds, i.e., when $\frac{a+b}{2} = \sqrt{ab}$.

Let's square both sides of the equality:

$\left(\frac{a+b}{2}\right)^2 = (\sqrt{ab})^2$

$\frac{(a+b)^2}{4} = ab$

Multiply both sides by 4:

$(a+b)^2 = 4ab$

Expand the left side:

$a^2 + 2ab + b^2 = 4ab$

Subtract $4ab$ from both sides:

$a^2 + 2ab + b^2 - 4ab = 0$

Combine the terms with $ab$:

$a^2 - 2ab + b^2 = 0$

The left side is a perfect square trinomial:

$(a-b)^2 = 0$

Take the square root of both sides:

$\sqrt{(a-b)^2} = \sqrt{0}$

$|a-b| = 0$

This implies:

$a-b = 0$

$a = b$

Thus, the equality in the AM-GM inequality for two positive numbers $a$ and $b$ holds if and only if $a$ is equal to $b$.

The correct option is (B) When $a=b$.

Solution:

Given:

Sum of the first 10 terms of an AP, $S_{10} = 210$.

Sum of the first 20 terms of an AP, $S_{20} = 820$.

To Find:

The first term ($a$) and the common difference ($d$) of the AP.

We use the formula for the sum of the first $n$ terms of an AP:

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

Using the given information for $n=10$:

$S_{10} = \frac{10}{2}[2a + (10-1)d]$

$210 = 5[2a + 9d]$

Divide both sides by 5:

$\frac{210}{5} = 2a + 9d$

$42 = 2a + 9d$

Let this be Equation (1).

Using the given information for $n=20$:

$S_{20} = \frac{20}{2}[2a + (20-1)d]$

$820 = 10[2a + 19d]$

Divide both sides by 10:

$\frac{820}{10} = 2a + 19d$

$82 = 2a + 19d$

Let this be Equation (2).

We now have a system of two linear equations:

(1) $2a + 9d = 42$

(2) $2a + 19d = 82$

Subtract Equation (1) from Equation (2) to eliminate $2a$:

$(2a + 19d) - (2a + 9d) = 82 - 42$

$2a + 19d - 2a - 9d = 40$

$10d = 40$

Divide both sides by 10:

$d = \frac{40}{10}$

$d = 4$

Substitute the value of $d=4$ into Equation (1):

$2a + 9d = 42$

$2a + 9(4) = 42$

$2a + 36 = 42$

Subtract 36 from both sides:

$2a = 42 - 36$

$2a = 6$

Divide both sides by 2:

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

$a = 3$

The first term is $a = 3$ and the common difference is $d = 4$.

Based on the calculations, the first term is 3 and the common difference is 4. This result does not match any of the provided options.

Solution:

Let $M$ be the arithmetic mean between two numbers $a$ and $b$.

If $M$ is the arithmetic mean between $a$ and $b$, then the sequence $a, M, b$ forms an Arithmetic Progression (AP).

In an AP, the difference between consecutive terms is constant.

Therefore, the difference between the second term ($M$) and the first term ($a$) must be equal to the difference between the third term ($b$) and the second term ($M$).

$M - a = b - M$

Now, we solve this equation for $M$. Add $M$ to both sides:

$M - a + M = b - M + M$

$2M - a = b$

Add $a$ to both sides:

$2M - a + a = b + a$

$2M = a + b$

Divide both sides by 2:

$M = \frac{a+b}{2}$

So, the arithmetic mean between $a$ and $b$ is $\frac{a+b}{2}$.

This corresponds to the formula for the Arithmetic Mean (AM) of two numbers.

Comparing this result with the given options, we find that option (C) matches the derived formula. Option (D) represents the Geometric Mean (GM), option (B) represents the product, and option (A) represents the sum.

The correct option is (C) $\frac{a+b}{2}$.

Solution:

Let the first term of the Geometric Progression (GP) be $a$ and the common ratio be $r$.

We are given the first term $a = 5$.

We are given the common ratio $r = -1$.

The terms of a GP are given by the sequence $a, ar, ar^2, ar^3, ar^4, ...$

Let's find the first few terms of the GP using the given values:

The first term is $a_1 = a = 5$.

The second term is $a_2 = ar = 5 \times (-1) = -5$.

The third term is $a_3 = ar^2 = 5 \times (-1)^2 = 5 \times 1 = 5$.

The fourth term is $a_4 = ar^3 = 5 \times (-1)^3 = 5 \times (-1) = -5$.

The fifth term is $a_5 = ar^4 = 5 \times (-1)^4 = 5 \times 1 = 5$.

The sequence of the terms of the GP is $5, -5, 5, -5, 5, ...$.

This sequence alternates between 5 and -5.

Comparing this sequence with the given options:

(A) 5, -5, 5, -5, ... (Alternating signs) - This matches our calculated sequence.

(B) 5, 5, 5, 5, ... (Constant sequence) - This would require a common ratio of 1.

(C) 5, 0, 0, 0, ... (After first term, rest are 0) - This would imply a common ratio of 0 (for terms after the first, if $a \ne 0$).

(D) 5, -5, -15, -25, ... (AP with common difference -10) - This is an Arithmetic Progression, not a Geometric Progression, as the differences are constant ($-5-5 = -10$, $-15-(-5) = -10$, etc.).

The sequence obtained from the given first term and common ratio matches option (A).

The correct option is (A) 5, -5, 5, -5, ... (Alternating signs).

Solution:

We are given an infinite Geometric Progression (GP) with:

First term, $a = 10$.

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

The sum of an infinite GP exists and converges to a finite value if and only if the absolute value of the common ratio is less than 1, i.e., $|r| < 1$.

In this case, the common ratio is $r = \frac{1}{2}$.

The absolute value is $|r| = \left|\frac{1}{2}\right| = \frac{1}{2}$.

Since $\frac{1}{2} < 1$, the sum of the infinite GP converges.

The formula for the sum of a convergent infinite GP is given by:

$S_\infty = \frac{a}{1-r}$

Substitute the given values of $a$ and $r$ into the formula:

$S_\infty = \frac{10}{1 - \frac{1}{2}}$

Calculate the denominator:

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

Now, substitute this back into the formula for $S_\infty$:

$S_\infty = \frac{10}{\frac{1}{2}}$

Dividing by a fraction is equivalent to multiplying by its reciprocal:

$S_\infty = 10 \times \frac{2}{1}$

$S_\infty = 20$

The sum of the infinite GP is 20.

The correct option is (B) 20.

Solution:

The given series is $1 + 11 + 111 + ...$.

The $n^{th}$ term of this series, denoted by $T_n$, is a number consisting of $n$ digits, all of which are 1.

We can express the $n^{th}$ term using powers of 10:

$T_n = 11...1$ ($n$ times)

Consider the identity $99...9$ ($n$ times) $= 10^n - 1$.

Since $T_n = \frac{1}{9} \times (99...9 \text{ n times})$, we have:

$T_n = \frac{1}{9} (10^n - 1)$

The sum of the first $n$ terms of the series, $S_n$, is $\sum_{k=1}^{n} T_k$:

$S_n = \sum_{k=1}^{n} \frac{1}{9} (10^k - 1)$

$S_n = \frac{1}{9} \sum_{k=1}^{n} (10^k - 1)$

Using the linearity property of summation, we can split the sum:

$S_n = \frac{1}{9} \left( \sum_{k=1}^{n} 10^k - \sum_{k=1}^{n} 1 \right)$

The first summation $\sum_{k=1}^{n} 10^k = 10^1 + 10^2 + ... + 10^n$ is the sum of a Geometric Progression with the first term $a = 10$ and the common ratio $r = 10$. The sum of the first $n$ terms of a GP is given by $S_n = a \frac{(r^n - 1)}{r - 1}$.

$\sum_{k=1}^{n} 10^k = 10 \frac{(10^n - 1)}{10 - 1} = 10 \frac{10^n - 1}{9}$

The second summation $\sum_{k=1}^{n} 1 = 1 + 1 + ... + 1$ ($n$ times) is simply $n$.

$\sum_{k=1}^{n} 1 = n$

Substitute these sums back into the expression for $S_n$:

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

This form matches option (A). Let's simplify this expression further to see if it matches option (D).

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

To combine the terms inside the parenthesis, find a common denominator, which is 9:

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

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

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

Multiply the fractions:

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

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

This simplified form matches option (D). Both options (A) and (D) are equivalent correct expressions for the sum of the series. Option (D) presents the sum in a more simplified, single-fraction form.

The sum of the first $n$ terms of the series $1 + 11 + 111 + ...$ is $\frac{1}{81} (10^{n+1} - 9n - 10)$.

The correct option is (D) $\frac{1}{81} (10^{n+1} - 9n - 10)$.

Solution:

We are given that $a, b, c$ are in Arithmetic Progression (AP).

If three numbers are in AP, the difference between consecutive terms is constant. Let the common difference be $d$.

The condition for $a, b, c$ to be in AP is:

$b - a = d$

and

$c - b = d$

From the first equation, we can write $a - b = -(b - a) = -d$.

Now, we need to find $(a-b)^2$.

$(a-b)^2 = (-d)^2 = d^2$.

Let's examine the given options:

(A) $(b-c)^2$

From the condition for AP, $b - c = -(c - b) = -d$.

So, $(b-c)^2 = (-d)^2 = d^2$.

This matches the value of $(a-b)^2$.

(B) $a^2 - c^2$

Using $b = a+d$ and $c = b+d = a+2d$:

$a^2 - c^2 = a^2 - (a+2d)^2 = a^2 - (a^2 + 4ad + 4d^2) = a^2 - a^2 - 4ad - 4d^2 = -4ad - 4d^2$.

This is not equal to $d^2$ in general.

(C) $(a-c)^2$

From the AP condition, $c - a = (c - b) + (b - a) = d + d = 2d$.

So, $a - c = -(c - a) = -2d$.

Then $(a-c)^2 = (-2d)^2 = 4d^2$.

This is not equal to $d^2$ in general (unless $d=0$).

(D) $(a+c)^2$

From the AP condition, $a+c = 2b$ (the middle term is the average of the other two).

So, $(a+c)^2 = (2b)^2 = 4b^2$.

This is not equal to $d^2$ in general.

Based on the analysis, $(a-b)^2$ is equal to $(b-c)^2$. Both are equal to the square of the common difference $d^2$.

The correct option is (A) $(b-c)^2$ (Common difference squared).

Solution:

A Geometric Progression (GP) is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$).

Let the sequence be $a_1, a_2, a_3, ..., a_n, ...$.

Based on the definition of a GP, the following conditions hold:

1. The ratio of any term to its preceding term is constant: $\frac{a_2}{a_1} = \frac{a_3}{a_2} = ... = \frac{a_n}{a_{n-1}} = r$, where $r$ is the common ratio. This means the ratio of consecutive terms is constant.

2. Each term (after the first) is obtained by multiplying the previous term by the common ratio: $a_n = a_{n-1} \times r$ for $n > 1$. This means each term (after the first) is the previous term multiplied by a constant ratio.

3. If the first term is $a$ and the common ratio is $r$, the terms of the GP can be written as $a, ar, ar^2, ar^3, ...$. The $n^{th}$ term is $a_n = ar^{n-1}$. This means the terms can be represented as $a, ar, ar^2, ...$.

Let's examine the given options in light of these properties:

(A) The ratio of consecutive terms is constant. This is a defining property of a GP. So, this IS a criterion.

(B) The difference of consecutive terms is constant. This is the defining property of an Arithmetic Progression (AP), not a GP (unless the common ratio of the GP is 1, resulting in a constant sequence, which is a special case of both AP and GP, but the general criterion for a GP is not the constant difference). So, this IS NOT a criterion for a GP in general.

(C) Each term (after the first) is the previous term multiplied by a constant ratio. This is a direct consequence of the definition of a GP ($a_n = a_{n-1} \cdot r$). So, this IS a criterion.

(D) The terms can be represented as $a, ar, ar^2, ...$. This is the standard form of representing terms in a GP. So, this IS a criterion.

The criterion that does NOT apply to a general Geometric Progression is that the difference of consecutive terms is constant.

The correct option, representing a statement that is NOT a criterion for a sequence to be a GP, is (B) The difference of consecutive terms is constant.

Solution:

The given series is $1 + 2 + 4 + 8 + ...$.

This is a Geometric Progression (GP) because the ratio of consecutive terms is constant.

The first term is $a = 1$.

The common ratio is $r = \frac{2}{1} = \frac{4}{2} = \frac{8}{4} = 2$.

The sum of the first $n$ terms of a GP is given by the formula:

$S_n = a \frac{(r^n - 1)}{r - 1}$, for $r \neq 1$.

We are given that the sum of the first $n$ terms is 127, so $S_n = 127$.

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

$127 = 1 \times \frac{(2^n - 1)}{2 - 1}$

$127 = \frac{2^n - 1}{1}$

$127 = 2^n - 1$

Add 1 to both sides of the equation:

$127 + 1 = 2^n$

$128 = 2^n$

Now, we need to find the power to which 2 must be raised to get 128. We can find this by repeatedly multiplying 2:

$2^1 = 2$

$2^2 = 4$

$2^3 = 8$

$2^4 = 16$

$2^5 = 32$

$2^6 = 64$

$2^7 = 128$

So, $2^n = 2^7$.

Since the bases are the same, the exponents must be equal.

$n = 7$

The value of $n$ is 7.

The correct option is (B) 7.

Solution:

Given:

The $n^{th}$ term of an AP is $a_n = 3n + 1$.

To Find:

The sum of the first 10 terms, $S_{10}$.

We can find the first term ($a_1$) by setting $n=1$ in the formula for $a_n$:

$a_1 = 3(1) + 1 = 3 + 1 = 4$

We can find the $10^{th}$ term ($a_{10}$) by setting $n=10$ in the formula for $a_n$:

$a_{10} = 3(10) + 1 = 30 + 1 = 31$

The sum of the first $n$ terms of an AP can be found using the formula:

$S_n = \frac{n}{2}(a_1 + a_n)$

For the sum of the first 10 terms, $n=10$:

$S_{10} = \frac{10}{2}(a_1 + a_{10})$

Substitute the values of $n$, $a_1$, and $a_{10}$:

$S_{10} = \frac{10}{2}(4 + 31)$

$S_{10} = 5(35)$

$S_{10} = 175$

Alternatively, we can find the common difference ($d$) of the AP. The common difference is the difference between any term and its preceding term.

$d = a_n - a_{n-1}$

We have $a_n = 3n + 1$.

To find $a_{n-1}$, substitute $(n-1)$ for $n$:

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

Now find the common difference:

$d = (3n + 1) - (3n - 2)$

$d = 3n + 1 - 3n + 2$

$d = 3$

The first term is $a_1 = 4$ and the common difference is $d = 3$.

The sum of the first $n$ terms of an AP can also be found using the formula:

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

For the sum of the first 10 terms, $n=10$:

$S_{10} = \frac{10}{2}[2a_1 + (10-1)d]$

Substitute the values of $n$, $a_1$, and $d$:

$S_{10} = 5[2(4) + 9(3)]$

$S_{10} = 5[8 + 27]$

$S_{10} = 5[35]$

$S_{10} = 175$

Both methods of calculation yield the same result, 175.

Option (A) shows the calculation using the formula $S_n = \frac{n}{2}(a_1 + a_n)$.

Option (B) shows the calculation using the formula $S_n = \frac{n}{2}[2a_1 + (n-1)d]$.

Option (C) provides the final numerical result.

Option (D) correctly states that all of the above options demonstrate the calculation or result of the calculation.

The sum of the first 10 terms is 175.

The correct option is (D) All of the above demonstrate the calculation.

Solution:

In mathematics, a **sequence** is an ordered list of numbers, points, functions, or other elements. The elements are listed in a particular order.

For example, the sequence of positive even numbers is $2, 4, 6, 8, ...$.

A **series** is the sum of the terms of a sequence.

For example, the series associated with the sequence of positive even numbers is $2 + 4 + 6 + 8 + ...$.

The fundamental difference is that a sequence is a collection of terms listed in order, while a series is the result of adding those terms together.

Let's evaluate the given statements based on these definitions:

(A) Every sequence is a series. This is incorrect. A sequence is a list, not a sum.

(B) Every series is a sequence. This is incorrect. A series is a sum, which results in a single value (if convergent) or a conceptual sum (if divergent). It is not an ordered list of terms itself.

(C) A sequence is a list of numbers, while a series is the sum of the terms of a sequence. This statement accurately captures the definition and distinction between a sequence and a series.

(D) There is no difference between a sequence and a series. This is incorrect. They are distinct mathematical concepts, although they are closely related.

Therefore, the correct statement is that a sequence is a list of numbers, while a series is the sum of the terms of a sequence.

The correct option is (C) A sequence is a list of numbers, while a series is the sum of the terms of a sequence.

Solution:

We are given that $a, b, c$ are in Geometric Progression (GP).

For three non-zero numbers to be in GP, the ratio of consecutive terms is constant. This means $\frac{b}{a} = \frac{c}{b}$.

Cross-multiplying, we get the condition:

$b^2 = ac$

We need to determine the type of progression formed by $\log a, \log b, \log c$. Assume that $a, b, c$ are positive so that their logarithms are defined.

Take the logarithm of both sides of the equation $b^2 = ac$ (using any valid base for the logarithm):

$\log(b^2) = \log(ac)$

Using the logarithm property $\log(M^k) = k \log M$:

$2 \log b = \log(ac)$

Using the logarithm property $\log(MN) = \log M + \log N$:

$2 \log b = \log a + \log c$

This equation, $2 \log b = \log a + \log c$, is the condition for three numbers to be in Arithmetic Progression (AP). Specifically, for three numbers $x, y, z$ to be in AP, the condition is $2y = x + z$.

Here, if we let $x = \log a$, $y = \log b$, and $z = \log c$, the equation $2 \log b = \log a + \log c$ matches the AP condition $2y = x + z$.

Therefore, if $a, b, c$ are in GP, then $\log a, \log b, \log c$ are in AP.

The correct option is (A) AP.

Solution:

Given:

The sum of an infinite Geometric Progression (GP) is $S_\infty = 15$.

The sum of the squares of its terms is $45$.

To Find:

The first term ($a$) and the common ratio ($r$) of the GP.

The sum of an infinite GP with first term $a$ and common ratio $r$ ($|r| < 1$) is given by the formula:

$S_\infty = \frac{a}{1-r}$

According to the given information:

The terms of the GP are $a, ar, ar^2, ar^3, ...$

The squares of the terms form a new sequence: $a^2, (ar)^2, (ar^2)^2, (ar^3)^2, ...$

This sequence is $a^2, a^2r^2, a^2r^4, a^2r^6, ...$

This new sequence is also a GP with the first term $a' = a^2$ and the common ratio $r' = \frac{a^2r^2}{a^2} = r^2$.

Since $|r|<1$, we have $|r^2| = |r|^2 < 1$, so the sum of this infinite GP also converges.

The sum of the squares of the terms is given as 45:

$S'_\infty = \frac{a'}{1-r'} = \frac{a^2}{1-r^2}$

According to the given information:

We now have a system of two equations with two variables $a$ and $r$.

From Equation (1), we can express $a$ in terms of $r$:

$a = 15(1-r)$

Substitute this expression for $a$ into Equation (2):

$\frac{(15(1-r))^2}{1-r^2} = 45$

Expand the numerator and use the difference of squares formula $1-r^2 = (1-r)(1+r)$ in the denominator:

$\frac{15^2 (1-r)^2}{(1-r)(1+r)} = 45$

$\frac{225 (1-r)(1-r)}{(1-r)(1+r)} = 45$

Since $|r|<1$, $r \ne 1$, so $1-r \ne 0$. We can cancel $(1-r)$ from the numerator and denominator:

$\frac{225 (1-r)}{1+r} = 45$

Divide both sides by 45:

$\frac{\cancel{225}^{5} (1-r)}{\cancel{45}^{1}(1+r)} = 1$

$\frac{5 (1-r)}{1+r} = 1$

Multiply both sides by $(1+r)$:

$5(1-r) = 1+r$

$5 - 5r = 1 + r$

Rearrange the terms to solve for $r$. Add $5r$ to both sides and subtract 1 from both sides:

$5 - 1 = r + 5r$

$4 = 6r$

Divide by 6:

$r = \frac{4}{6} = \frac{2}{3}$

The common ratio is $r = \frac{2}{3}$. This satisfies the condition $|r| < 1$.

Now substitute the value of $r = \frac{2}{3}$ back into the expression for $a$ from Equation (1):

$a = 15(1-r)$

$a = 15\left(1 - \frac{2}{3}\right)$

$a = 15\left(\frac{3}{3} - \frac{2}{3}\right)$

$a = 15\left(\frac{1}{3}\right)$

$a = \frac{\cancel{15}^{5}}{\cancel{3}^{1}}$

$a = 5$

The first term is $a=5$ and the common ratio is $r=2/3$.

The first term is 5 and the common ratio is 2/3.

Comparing this result with the given options, we find that it matches option (A).

The correct option is (A) $a=5, r=2/3$.

Solution:

The sequence is defined by the formula:

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

We need to find the first five terms of the sequence, which means we need to calculate $a_1, a_2, a_3, a_4$, and $a_5$.

For the first term ($n=1$):

$a_1 = \frac{1(1+1)}{2} = \frac{1(2)}{2} = \frac{2}{2} = 1$

For the second term ($n=2$):

$a_2 = \frac{2(2+1)}{2} = \frac{2(3)}{2} = \frac{6}{2} = 3$

For the third term ($n=3$):

$a_3 = \frac{3(3+1)}{2} = \frac{3(4)}{2} = \frac{12}{2} = 6$

For the fourth term ($n=4$):

$a_4 = \frac{4(4+1)}{2} = \frac{4(5)}{2} = \frac{20}{2} = 10$

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

$a_5 = \frac{5(5+1)}{2} = \frac{5(6)}{2} = \frac{30}{2} = 15$

Thus, the first five terms of the sequence defined by $a_n = \frac{n(n+1)}{2}$ are:

1, 3, 6, 10, 15.

Solution:

Given:

First term of the Arithmetic Progression, $a = 5$.

Common difference, $d = 3$.

To Find:

The 15th term of the Arithmetic Progression, $a_{15}$.

Solution:

The formula for the n-th term of an Arithmetic Progression is given by:

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

Here, we need to find the 15th term, so we substitute $n=15$ into the formula.

We have $a = 5$, $d = 3$, and $n = 15$.

Substitute these values into the formula:

$a_{15} = 5 + (15-1) \times 3$

Calculate the value:

$a_{15} = 5 + (14) \times 3$

$a_{15} = 5 + 42$

$a_{15} = 47$

The 15th term of the Arithmetic Progression is 47.

Solution:

Given:

The 3rd term of an AP, $a_3 = 7$.

The 7th term of an AP, $a_7 = 19$.

To Find:

The common difference ($d$) and the first term ($a$) of the AP.

Solution:

The formula for the n-th term of an Arithmetic Progression is given by:

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

Using the given information, we can write two equations:

Substituting the value of $a_3$:

Substituting the value of $a_7$:

Now we have a system of two linear equations:

We can solve this system by subtracting Equation (1) from Equation (2):

Divide by 4 to find the value of $d$:

Now substitute the value of $d = 3$ into Equation (1) to find the value of $a$:

Subtract 6 from both sides:

The common difference is 3 and the first term is 1.

Solution:

Given:

The Arithmetic Progression (AP) is: 1, 5, 9, 13, ...

The number of terms to sum, $n = 20$.

To Find:

The sum of the first 20 terms of the AP, $S_{20}$.

Solution:

From the given AP, the first term is $a = 1$.

The common difference $d$ is the difference between consecutive terms:

$d = 5 - 1 = 4$

$d = 9 - 5 = 4$

So, the common difference is $d = 4$.

The formula for the sum of the first $n$ terms of an Arithmetic Progression is given by:

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

We need to find the sum of the first 20 terms, so we substitute $n=20$, $a=1$, and $d=4$ into the formula.

$S_{20} = \frac{20}{2}[2(1) + (20-1)(4)]$

Now, we calculate the value:

$S_{20} = 10[2 + (19)(4)]$

$S_{20} = 10[2 + 76]$

$S_{20} = 10[78]$

$S_{20} = 780$

The sum of the first 20 terms of the AP is 780.

Solution:

Given:

Two numbers, 10 and 34.

We need to insert two Arithmetic Means between them.

To Find:

The two Arithmetic Means between 10 and 34.

Solution:

Let the two Arithmetic Means between 10 and 34 be $A_1$ and $A_2$.

Then, the sequence 10, $A_1$, $A_2$, 34 forms an Arithmetic Progression.

In this AP:

The first term is $a_1 = 10$.

The number of terms is $n = 4$ (10, $A_1$, $A_2$, 34).

The fourth term is $a_4 = 34$.

The formula for the n-th term of an Arithmetic Progression is given by:

$a_n = a_1 + (n-1)d$

where $a_1$ is the first term and $d$ is the common difference.

Using the fourth term ($n=4$), we have:

$a_4 = a_1 + (4-1)d$

$34 = 10 + 3d$

Now, we solve for the common difference $d$:

$3d = 34 - 10$

$3d = 24$

$d = \frac{24}{3}$

$d = 8$

The two Arithmetic Means are $A_1 = a_2$ and $A_2 = a_3$.

$A_1 = a_1 + d = 10 + 8 = 18$

$A_2 = a_1 + 2d = 10 + 2(8) = 10 + 16 = 26$

The two Arithmetic Means between 10 and 34 are 18 and 26.

The AP is 10, 18, 26, 34.

Solution:

Given:

The Geometric Progression (GP) is: 3, 6, 12, ...

To Find:

The 8th term of the Geometric Progression, $a_8$.

Solution:

From the given GP, the first term is $a = 3$.

The common ratio $r$ is the ratio of consecutive terms:

$r = \frac{6}{3} = 2$

$r = \frac{12}{6} = 2$

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

The formula for the n-th term of a Geometric Progression is given by:

$a_n = ar^{n-1}$

We need to find the 8th term, so we substitute $n=8$, $a=3$, and $r=2$ into the formula.

$a_8 = 3 \times (2)^{8-1}$

$a_8 = 3 \times (2)^7$

Now, we calculate the value of $2^7$:

$2^1 = 2$

$2^2 = 4$

$2^3 = 8$

$2^4 = 16$

$2^5 = 32$

$2^6 = 64$

$2^7 = 128$

Substitute this back into the expression for $a_8$:

$a_8 = 3 \times 128$

Perform the multiplication:

$a_8 = 384$

The 8th term of the Geometric Progression is 384.

Solution:

Given:

The first term of a Geometric Progression, $a_1 = 2$.

The 5th term of the Geometric Progression, $a_5 = 162$.

To Find:

The common ratio ($r$) of the GP.

Solution:

The formula for the n-th term of a Geometric Progression is given by:

$a_n = a_1 r^{n-1}$

where $a_1$ is the first term and $r$ is the common ratio.

We are given the first term $a_1 = 2$.

We are also given the 5th term $a_5 = 162$. Using the formula for $n=5$:

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

Now, we need to solve Equation (1) for $r$.

Divide both sides of the equation by 2:

$\frac{162}{2} = r^4$

$81 = r^4$

To find $r$, we need to find the fourth root of 81.

$r = \sqrt[4]{81}$

We look for a number that, when multiplied by itself four times, equals 81.

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

Also, $(-3) \times (-3) \times (-3) \times (-3) = (-3)^4 = 81$.

So, the possible values for $r$ are $3$ and $-3$.

$r = 3$ or $r = -3$

The common ratio can be 3 or -3.

(Note: Without further information, both positive and negative common ratios are possible.)

Solution:

Given:

The Geometric Progression (GP) is: 1, $\frac{1}{2}$, $\frac{1}{4}$, ...

The number of terms to sum, $n = 6$.

To Find:

The sum of the first 6 terms of the GP, $S_6$.

Solution:

From the given GP, the first term is $a = 1$.

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

$r = \frac{\frac{1}{2}}{1} = \frac{1}{2}$

or

$r = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \times 2 = \frac{2}{4} = \frac{1}{2}$

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

The formula for the sum of the first $n$ terms of a Geometric Progression is given by:

$S_n = a \frac{(1 - r^n)}{(1 - r)}$, when $r \neq 1$.

We need to find the sum of the first 6 terms, so we substitute $n=6$, $a=1$, and $r=\frac{1}{2}$ into the formula.

$S_6 = 1 \times \frac{(1 - (\frac{1}{2})^6)}{(1 - \frac{1}{2})}$

First, calculate $(\frac{1}{2})^6$:

$(\frac{1}{2})^6 = \frac{1^6}{2^6} = \frac{1}{64}$

Substitute this value back into the formula for $S_6$:

$S_6 = 1 \times \frac{(1 - \frac{1}{64})}{(1 - \frac{1}{2})}$

Simplify the numerator and the denominator:

Numerator: $1 - \frac{1}{64} = \frac{64}{64} - \frac{1}{64} = \frac{63}{64}$

Denominator: $1 - \frac{1}{2} = \frac{2}{2} - \frac{1}{2} = \frac{1}{2}$

Now substitute these simplified values:

$S_6 = \frac{\frac{63}{64}}{\frac{1}{2}}$

To divide by a fraction, multiply by its reciprocal:

$S_6 = \frac{63}{64} \times \frac{2}{1}$

$S_6 = \frac{63 \times 2}{64 \times 1}$

$S_6 = \frac{126}{64}$

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

$S_6 = \frac{\cancel{126}^{63}}{\cancel{64}_{32}}$

$S_6 = \frac{63}{32}$

The sum of the first 6 terms of the GP is $\frac{63}{32}$.

Solution:

Given:

Two numbers, 5 and 80.

We need to insert one Geometric Mean between them.

To Find:

The Geometric Mean between 5 and 80.

Solution:

Let the Geometric Mean between two numbers $a$ and $b$ be $G$.

By definition, $a, G, b$ form a Geometric Progression (GP).

This means the ratio of consecutive terms is constant (the common ratio, $r$).

So, $\frac{G}{a} = \frac{b}{G}$.

Cross-multiplying gives:

$G^2 = ab$

Therefore, the Geometric Mean is $G = \pm\sqrt{ab}$.

In this problem, $a = 5$ and $b = 80$.

Substitute these values into the formula:

$G = \pm\sqrt{5 \times 80}$

$G = \pm\sqrt{400}$

We calculate the square root of 400.

$\sqrt{400} = 20$ since $20 \times 20 = 400$.

So, the possible values for the Geometric Mean are:

$G = 20$ or $G = -20$

If $G=20$, the sequence is 5, 20, 80, which is a GP with common ratio $\frac{20}{5} = 4$ and $\frac{80}{20} = 4$.

If $G=-20$, the sequence is 5, -20, 80, which is a GP with common ratio $\frac{-20}{5} = -4$ and $\frac{80}{-20} = -4$.

The Geometric Mean between 5 and 80 can be 20 or -20.

Solution:

Given:

The Arithmetic Mean (AM) of two positive numbers is 10.

The Geometric Mean (GM) of the same two positive numbers is 8.

To Find:

The two positive numbers.

Solution:

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

The formula for the Arithmetic Mean of two numbers $a$ and $b$ is $\frac{a+b}{2}$.

The formula for the Geometric Mean of two positive numbers $a$ and $b$ is $\sqrt{ab}$.

According to the problem statement, we have the following equations:

From Equation (1), we can find the sum of the two numbers:

From Equation (2), we can find the product of the two numbers by squaring both sides:

Now we have a system of two equations:

$a+b = 20$

$ab = 64$

From Equation (3), express $b$ in terms of $a$: $b = 20 - a$.

Substitute this expression for $b$ into Equation (4):

Expand the equation:

Rearrange into a standard quadratic form ($ax^2 + bx + c = 0$):

We can solve this quadratic equation by factoring. We look for two numbers that multiply to 64 and add up to -20. These numbers are -4 and -16.

So, we can factor the equation as:

This equation is satisfied if either factor is zero.

Case 1: $a - 4 = 0 \implies a = 4$.

Substitute $a=4$ into Equation (3) ($a+b=20$): $4 + b = 20 \implies b = 16$.

Case 2: $a - 16 = 0 \implies a = 16$.

Substitute $a=16$ into Equation (3) ($a+b=20$): $16 + b = 20 \implies b = 4$.

In both cases, the two numbers are 4 and 16.

Since the problem specifies "two positive numbers", both 4 and 16 are valid.

The two numbers are 4 and 16.

Verification:

AM of 4 and 16 = $\frac{4+16}{2} = \frac{20}{2} = 10$ (Correct).

GM of 4 and 16 = $\sqrt{4 \times 16} = \sqrt{64} = 8$ (Correct).

Alternate Method (Using a standard identity):

If $a$ and $b$ are two numbers, their AM is $A = \frac{a+b}{2}$ and GM is $G = \sqrt{ab}$.

The quadratic equation whose roots are $a$ and $b$ is given by $x^2 - (a+b)x + ab = 0$.

From the given information, $a+b = 2A = 2 \times 10 = 20$.

$ab = G^2 = 8^2 = 64$.

Substituting these values into the quadratic equation:

Solving this quadratic equation (as done above) gives the roots as $x = 4$ and $x = 16$.

Thus, the two numbers are 4 and 16.

The two numbers are 4 and 16.

Solution:

Given:

The series is the sum of the squares of the first 10 natural numbers: $1^2 + 2^2 + 3^2 + \dots + 10^2$.

To Find:

The sum of the series.

Solution:

The sum of the squares of the first $n$ natural numbers is given by the formula:

In this problem, we need to find the sum of the first 10 terms, so $n = 10$.

Substitute $n=10$ into the formula:

$S_{10} = \frac{10(10+1)(2 \times 10 + 1)}{6}$

$S_{10} = \frac{10(11)(20 + 1)}{6}$

$S_{10} = \frac{10 \times 11 \times 21}{6}$

Now, perform the calculation. We can simplify the fraction before multiplying.

Divide 10 by 2 and 6 by 2:

$S_{10} = \frac{\cancel{10}^{5} \times 11 \times 21}{\cancel{6}_{3}}$

Divide 21 by 3 and 3 by 3:

$S_{10} = \frac{5 \times 11 \times \cancel{21}^{7}}{\cancel{3}_{1}}$

$S_{10} = 5 \times 11 \times 7$

Multiply the remaining numbers:

$S_{10} = 55 \times 7$

$S_{10} = 385$

The sum of the series $1^2 + 2^2 + \dots + 10^2$ is 385.

Solution:

Given:

The $n$th term of the sequence is given by $a_n = 3n - 2$.

To Show:

That the sequence is an Arithmetic Progression (AP).

To Find:

The first term and the common difference of the AP.

Solution:

To show that a sequence is an Arithmetic Progression, we need to demonstrate that the difference between any term and its preceding term is a constant value. This constant difference is the common difference ($d$).

The $n$th term of the sequence is $a_n = 3n - 2$.

The $(n+1)$th term of the sequence is found by replacing $n$ with $(n+1)$:

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

$a_{n+1} = 3n + 3 - 2$

$a_{n+1} = 3n + 1$

Now, let's find the difference between the $(n+1)$th term and the $n$th term:

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

$a_{n+1} - a_n = 3n + 1 - 3n + 2$

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

$a_{n+1} - a_n = 0 + 3$

$a_{n+1} - a_n = 3$

Since the difference between any term and its preceding term ($a_{n+1} - a_n$) is a constant value (which is 3) for all $n \geq 1$, the given sequence is an Arithmetic Progression.

The common difference is this constant value, so the common difference is $d = 3$.

To find the first term of the sequence, we substitute $n=1$ into the formula for $a_n$:

$a_1 = 3(1) - 2$

$a_1 = 3 - 2$

$a_1 = 1$

The first term of the sequence is $a_1 = 1$.

Solution:

Given:

The Arithmetic Progression (AP) is: 2, 7, 12, 17, ...

A specific term value is 87.

To Find:

The term number ($n$) of the AP that is equal to 87.

Solution:

From the given AP, the first term is $a = 2$.

The common difference $d$ is the difference between consecutive terms:

$d = 7 - 2 = 5$

$d = 12 - 7 = 5$

So, the common difference is $d = 5$.

We want to find which term is 87. Let the $n$th term be $a_n = 87$.

The formula for the n-th term of an Arithmetic Progression is given by:

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

Substitute the known values into the formula:

$87 = 2 + (n-1)5$

Now, we solve this equation for $n$:

Subtract 2 from both sides:

$87 - 2 = (n-1)5$

$85 = (n-1)5$

Divide both sides by 5:

Perform the division:

$17 = n-1$

Add 1 to both sides to find $n$:

$17 + 1 = n$

$n = 18$

The 18th term of the AP is 87.

Solution:

Given:

The Arithmetic Progression (AP) is: 24, 21, 18, ...

The sum of $n$ terms is $S_n = 78$.

To Find:

The number of terms ($n$) that must be taken so that their sum is 78.

Solution:

From the given AP, we can identify:

The first term, $a = 24$.

The common difference, $d = 21 - 24 = -3$.

The formula for the sum of the first $n$ terms of an Arithmetic Progression is given by:

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

We are given that the sum $S_n = 78$. Substitute the values of $S_n$, $a$, and $d$ into the formula:

Now, we need to solve this equation for $n$.

Multiply both sides by 2:

$78 \times 2 = n[48 - 3(n-1)]$

$156 = n[48 - 3n + 3]$

$156 = n[51 - 3n]$

$156 = 51n - 3n^2$

Rearrange the equation into a standard quadratic form $Ax^2 + Bx + C = 0$:

$3n^2 - 51n + 156 = 0$

Divide the entire equation by 3 to simplify it:

$\frac{3n^2}{3} - \frac{51n}{3} + \frac{156}{3} = 0$

We can solve this quadratic equation by factoring. We look for two numbers that multiply to 52 and add up to -17. These numbers are -4 and -13.

So, we can factor the quadratic expression as:

$(n - 4)(n - 13) = 0$

For the product of two factors to be zero, at least one of the factors must be zero.

Case 1:

$n - 4 = 0$

$n = 4$

Case 2:

$n - 13 = 0$

$n = 13$

Both $n=4$ and $n=13$ are positive integers and thus valid numbers of terms.

This means that the sum of the first 4 terms of the AP is 78, and the sum of the first 13 terms of the AP is also 78.

This occurs because the terms of the AP are decreasing (due to a negative common difference), and the terms from the 5th term onwards sum to zero ($a_5 + \dots + a_{13} = 0$).

The number of terms that must be taken for their sum to be 78 are 4 or 13.

Solution:

Given:

Two numbers are 8 and 26.

We need to insert 5 Arithmetic Means between them.

To Find:

The 5 Arithmetic Means between 8 and 26.

Solution:

Let the 5 Arithmetic Means between 8 and 26 be $A_1, A_2, A_3, A_4, A_5$.

When these means are inserted, the sequence 8, $A_1, A_2, A_3, A_4, A_5$, 26 forms an Arithmetic Progression (AP).

In this AP:

The first term is $a_1 = 8$.

The last term is $a_n = 26$.

The total number of terms in this AP is the two original numbers plus the 5 inserted means, so $n = 2 + 5 = 7$.

Thus, the 7th term of the AP is $a_7 = 26$.

The formula for the n-th term of an Arithmetic Progression is given by:

$a_n = a_1 + (n-1)d$

where $a_1$ is the first term and $d$ is the common difference.

Using the 7th term ($n=7$), we have:

Substitute the known values ($a_7=26$, $a_1=8$) into the equation:

Now, we solve Equation (1) for the common difference $d$:

Subtract 8 from both sides:

Divide both sides by 6:

Now that we have the common difference ($d=3$) and the first term ($a_1=8$), we can find the 5 Arithmetic Means. The means are the terms $a_2, a_3, a_4, a_5, a_6$ of the AP.

First Arithmetic Mean ($A_1 = a_2$):

$A_1 = a_1 + d = 8 + 3 = 11$

Second Arithmetic Mean ($A_2 = a_3$):

$A_2 = a_1 + 2d = 8 + 2(3) = 8 + 6 = 14$

Third Arithmetic Mean ($A_3 = a_4$):

$A_3 = a_1 + 3d = 8 + 3(3) = 8 + 9 = 17$

Fourth Arithmetic Mean ($A_4 = a_5$):

$A_4 = a_1 + 4d = 8 + 4(3) = 8 + 12 = 20$

Fifth Arithmetic Mean ($A_5 = a_6$):

$A_5 = a_1 + 5d = 8 + 5(3) = 8 + 15 = 23$

The 5 Arithmetic Means between 8 and 26 are 11, 14, 17, 20, and 23.

The resulting AP is 8, 11, 14, 17, 20, 23, 26.

Solution:

Given:

The 4th term of a Geometric Progression (GP), $a_4 = 24$.

The 7th term of the GP, $a_7 = 192$.

To Find:

The first term ($a$) and the common ratio ($r$) of the GP.

Solution:

The formula for the n-th term of a Geometric Progression is given by:

$a_n = ar^{n-1}$

where $a$ is the first term and $r$ is the common ratio.

Using the given information, we can write two equations:

Now we have a system of two equations with two variables ($a$ and $r$):

$ar^3 = 24$

$ar^6 = 192$

To find the common ratio $r$, we can divide Equation (2) by Equation (1):

Simplify both sides:

To find $r$, take the cube root of both sides of Equation (3):

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

Since $2 \times 2 \times 2 = 8$, the real value of $r$ is 2.

Now that we have the common ratio ($r=2$), we can substitute this value into Equation (1) to find the first term $a$:

Calculate $2^3$:

$2^3 = 2 \times 2 \times 2 = 8$

Substitute this value back into the equation:

Solve for $a$ by dividing both sides by 8:

The first term of the GP is 3 and the common ratio is 2.

Verification:

First term = 3.

Common ratio = 2.

4th term = $a r^{4-1} = 3 \times 2^3 = 3 \times 8 = 24$ (Correct).

7th term = $a r^{7-1} = 3 \times 2^6 = 3 \times 64 = 192$ (Correct).

Solution:

Given:

The Geometric Progression (GP) is: 0.15, 0.015, 0.0015, ...

The number of terms to sum, $n = 8$.

To Find:

The sum of the first 8 terms of the GP, $S_8$.

Solution:

From the given GP, we can identify the first term and the common ratio.

The first term is $a = 0.15$.

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

$r = \frac{0.015}{0.15}$

To simplify this division, we can multiply the numerator and denominator by 1000:

$r = \frac{0.015 \times 1000}{0.15 \times 1000} = \frac{15}{150} = \frac{1}{10}$

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

Since the common ratio $|r| = |0.1| = 0.1 < 1$, the formula for the sum of the first $n$ terms of a Geometric Progression is given by:

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

We need to find the sum of the first 8 terms, so we substitute $n=8$, $a=0.15$, and $r=0.1$ into the formula.

First, calculate $(0.1)^8$:

$(0.1)^8 = (10^{-1})^8 = 10^{-8} = 0.00000001$

Next, calculate the terms in the denominator and numerator of the fraction:

$1 - r = 1 - 0.1 = 0.9$

$1 - r^n = 1 - (0.1)^8 = 1 - 0.00000001 = 0.99999999$

Substitute these values back into the formula for $S_8$ (Equation i):

$S_8 = 0.15 \times \frac{0.99999999}{0.9}$

We can calculate $\frac{0.15}{0.9}$ first:

$\frac{0.15}{0.9} = \frac{15}{90} = \frac{1}{6}$

So, the sum becomes:

$S_8 = \frac{1}{6} \times 0.99999999$

$S_8 = \frac{0.99999999}{6}$

Now, perform the division:

$0.99999999 \div 6$

$S_8 = 0.166666665$

Alternatively, using fractions:

$a = \frac{15}{100} = \frac{3}{20}$

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

$r^8 = (\frac{1}{10})^8 = \frac{1}{10^8}$

$1 - r^8 = 1 - \frac{1}{10^8} = \frac{10^8 - 1}{10^8}$

$1 - r = 1 - \frac{1}{10} = \frac{9}{10}$

$S_8 = \frac{a(1-r^8)}{1-r} = \frac{\frac{3}{20} \times (\frac{10^8 - 1}{10^8})}{\frac{9}{10}}$

$S_8 = \frac{3}{20} \times \frac{10^8 - 1}{10^8} \times \frac{10}{9}$

$S_8 = \frac{3 \times 10 \times (10^8 - 1)}{20 \times 10^8 \times 9}$

$S_8 = \frac{\cancel{30}^{1} \times (10^8 - 1)}{\cancel{180}_{6} \times 10^8}$

$S_8 = \frac{10^8 - 1}{6 \times 10^8}$

$10^8 = 100,000,000$

$S_8 = \frac{100,000,000 - 1}{6 \times 100,000,000} = \frac{99,999,999}{600,000,000}$

Dividing numerator and denominator by 3:

$S_8 = \frac{33,333,333}{200,000,000}$

This fraction is equal to the decimal 0.166666665.

The sum of the first 8 terms of the GP is 0.166666665 or $\frac{33333333}{200000000}$.

Solution:

Given:

The Arithmetic Mean (AM) of two positive numbers $a$ and $b$ is 15.

The Geometric Mean (GM) of the same two positive numbers $a$ and $b$ is 9.

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 two numbers $a$ and $b$ is $\frac{a+b}{2}$.

The formula for the Geometric Mean of two positive numbers $a$ and $b$ is $\sqrt{ab}$.

According to the problem statement, we have the following equations:

From Equation (1), we can find the sum of the two numbers:

From Equation (2), we can find the product of the two numbers by squaring both sides:

Now we have a system of two equations:

$a+b = 30$

$ab = 81$

From Equation (3), express $b$ in terms of $a$: $b = 30 - a$.

Substitute this expression for $b$ into Equation (4):

Expand the equation:

Rearrange into a standard quadratic form ($Ax^2 + Bx + C = 0$):

We can solve this quadratic equation by factoring. We look for two numbers that multiply to 81 and add up to -30. These numbers are -3 and -27.

So, we can factor the equation as:

This equation is satisfied if either factor is zero.

Case 1: $a - 3 = 0 \implies a = 3$.

Substitute $a=3$ into Equation (3) ($a+b=30$): $3 + b = 30 \implies b = 27$.

Case 2: $a - 27 = 0 \implies a = 27$.

Substitute $a=27$ into Equation (3) ($a+b=30$): $27 + b = 30 \implies b = 3$.

In both cases, the two numbers are 3 and 27.

Since the problem specifies "two positive numbers", both 3 and 27 are valid.

The two positive numbers are 3 and 27.

Verification:

AM of 3 and 27 = $\frac{3+27}{2} = \frac{30}{2} = 15$ (Correct).

GM of 3 and 27 = $\sqrt{3 \times 27} = \sqrt{81} = 9$ (Correct).

Alternate Method (Using a standard identity):

If $a$ and $b$ are two numbers, their AM is $A = \frac{a+b}{2}$ and GM is $G = \sqrt{ab}$.

The quadratic equation whose roots are $a$ and $b$ is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.

From the given information, the sum of the numbers is $a+b = 2 \times AM = 2 \times 15 = 30$.

The product of the numbers is $ab = GM^2 = 9^2 = 81$.

Substituting these values into the quadratic equation:

We solve this quadratic equation for $x$. We are looking for values of $x$ that satisfy the equation. We can factor it as $(x-3)(x-27) = 0$.

This yields the roots $x = 3$ and $x = 27$.

Thus, the two numbers are 3 and 27.

The two positive numbers are 3 and 27.

Solution:

Given:

The series is given by the summation $\sum\limits_{k=1}^{10} (2k^2 - 3k + 5)$.

To Find:

The sum of the series.

Solution:

We need to find the sum of the terms $(2k^2 - 3k + 5)$ for $k$ from 1 to 10.

We can use the properties of summation, which allow us to split the sum and pull out constants:

$\sum\limits_{k=1}^{10} (2k^2 - 3k + 5) = \sum\limits_{k=1}^{10} (2k^2) - \sum\limits_{k=1}^{10} (3k) + \sum\limits_{k=1}^{10} (5)$

= $2 \sum\limits_{k=1}^{10} k^2 - 3 \sum\limits_{k=1}^{10} k + \sum\limits_{k=1}^{10} 5$

We use the standard formulas for the sum of the first $n$ natural numbers and the sum of the squares of the first $n$ natural numbers:

1. Sum of a constant: $\sum\limits_{k=1}^{n} c = nc$

2. Sum of the first $n$ natural numbers: $\sum\limits_{k=1}^{n} k = \frac{n(n+1)}{2}$

3. Sum of the squares of the first $n$ natural numbers: $\sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

In this problem, $n=10$. Let's calculate each part of the sum for $n=10$.

First part: $\sum\limits_{k=1}^{10} 5$

Using formula 1 with $c=5$ and $n=10$:

$\sum\limits_{k=1}^{10} 5 = 5 \times 10 = 50$

Second part: $- 3 \sum\limits_{k=1}^{10} k$

Using formula 2 with $n=10$:

$\sum\limits_{k=1}^{10} k = \frac{10(10+1)}{2} = \frac{10 \times 11}{2} = \frac{110}{2} = 55$

So, $- 3 \sum\limits_{k=1}^{10} k = -3 \times 55$

$= -165$

Third part: $2 \sum\limits_{k=1}^{10} k^2$

Using formula 3 with $n=10$:

$\sum\limits_{k=1}^{10} k^2 = \frac{10(10+1)(2 \times 10 + 1)}{6} = \frac{10(11)(21)}{6}$

$\sum\limits_{k=1}^{10} k^2 = \frac{10 \times 11 \times 21}{6} = \frac{\cancel{10}^{5} \times 11 \times \cancel{21}^{7}}{\cancel{6}_{3}} = 5 \times 11 \times 7 = 55 \times 7 = 385$

So, $2 \sum\limits_{k=1}^{10} k^2 = 2 \times 385$

$= 770$

Now, combine the results of the three parts:

Sum $= 770 - 165 + 50$

Sum $= (770 + 50) - 165$

Sum $= 820 - 165$

Perform the subtraction:

Sum $= 655$

The sum of the series $\sum\limits_{k=1}^{10} (2k^2 - 3k + 5)$ is 655.

Solution:

Given:

The sequence is defined by:

To Find:

The first four terms of the sequence.

Solution:

We are given the first term directly.

First term ($n=1$):

Second term ($n=2$):

Using the recursive relation $a_n = 3a_{n-1} + 2$ for $n=2$:

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

$a_2 = 3a_1 + 2$

Substitute the value of $a_1 = 3$:

$a_2 = 3(3) + 2$

$a_2 = 9 + 2$

$a_2 = 11$

Third term ($n=3$):

Using the recursive relation $a_n = 3a_{n-1} + 2$ for $n=3$:

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

$a_3 = 3a_2 + 2$

Substitute the value of $a_2 = 11$:

$a_3 = 3(11) + 2$

$a_3 = 33 + 2$

$a_3 = 35$

Fourth term ($n=4$):

Using the recursive relation $a_n = 3a_{n-1} + 2$ for $n=4$:

$a_4 = 3a_{4-1} + 2$

$a_4 = 3a_3 + 2$

Substitute the value of $a_3 = 35$:

$a_4 = 3(35) + 2$

$a_4 = 105 + 2$

$a_4 = 107$

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

Solution:

Given:

In an Arithmetic Progression (AP), the $m$th term is denoted by $a_m$ and the $n$th term by $a_n$.

The given condition is that $m$ times the $m$th term is equal to $n$ times the $n$th term, which can be written as:

We are also given that $m \neq n$.

To Prove:

The $(m+n)$th term of the AP is zero, i.e., $a_{m+n} = 0$.

Proof:

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

The formula for the $k$th term of an AP is given by:

$a_k = a + (k-1)d$

Using this formula, we can express the $m$th term and the $n$th term as:

Now, substitute these expressions for $a_m$ and $a_n$ into the given condition $m \times a_m = n \times a_n$:

Expand both sides of the equation:

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

$ma + (m^2 - m)d = na + (n^2 - n)d$

Gather the terms containing $a$ on one side and the terms containing $d$ on the other side:

$ma - na = (n^2 - n)d - (m^2 - m)d$

$a(m - n) = (n^2 - n - m^2 + m)d$

Rearrange the terms on the right side:

$a(m - n) = (n^2 - m^2 + m - n)d$

Factor the difference of squares ($n^2 - m^2 = (n-m)(n+m)$) and group the last two terms:

$a(m - n) = [(n-m)(n+m) + (m - n)]d$

We can rewrite $(n-m)$ as $-(m-n)$. Substitute this into the equation:

$a(m - n) = [-(m-n)(n+m) + (m - n)]d$

Now, factor out the common term $(m - n)$ from the expression in the square brackets on the right side:

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

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

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

Since it is given that $m \neq n$, we know that the term $(m - n)$ is not equal to zero. Therefore, we can divide both sides of the equation by $(m - n)$:

$\frac{a(m - n)}{(m - n)} = \frac{(m - n)(1 - n - m)d}{(m - n)}$

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

Rearrange the terms to group $a$ and the term involving $d$:

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

Add $(m+n-1)d$ to both sides of the equation:

Now, let's write the expression for the $(m+n)$th term of the AP using the formula $a_k = a + (k-1)d$ with $k = m+n$:

Compare Equation (3) and Equation (4). We see that the expression for $a_{m+n}$ is exactly equal to the expression that we proved is equal to zero.

Substituting the result from Equation (3) into Equation (4):

$a_{m+n} = 0$

Thus, the $(m+n)$th term of the Arithmetic Progression is zero.

This completes the proof.

Solution:

Given:

The infinite geometric series is: $1 + \frac{1}{3} + \frac{1}{9} + \dots$

To Find:

The sum of the infinite geometric series.

Solution:

The given series is a Geometric Progression (GP).

The first term is $a = 1$.

The common ratio $r$ is found by dividing the second term by the first term, or the third term by the second term:

$r = \frac{\frac{1}{3}}{1} = \frac{1}{3}$

or

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

So, the common ratio is $r = \frac{1}{3}$.

For an infinite geometric series to have a finite sum (to converge), the absolute value of the common ratio must be less than 1, i.e., $|r| < 1$.

In this case, $|r| = |\frac{1}{3}| = \frac{1}{3}$.

Since $\frac{1}{3} < 1$, the sum of the infinite geometric series converges.

The formula for the sum of an infinite Geometric Progression with first term $a$ and common ratio $r$ (where $|r| < 1$) is given by:

$S_\infty = \frac{a}{1-r}$

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

$S_\infty = \frac{1}{1 - \frac{1}{3}}$

Calculate the value in the denominator:

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

Substitute this back into the formula for $S_\infty$:

$S_\infty = \frac{1}{\frac{2}{3}}$

To divide by a fraction, multiply by its reciprocal:

$S_\infty = 1 \times \frac{3}{2}$

$S_\infty = \frac{3}{2}$

The sum of the infinite geometric series is $\frac{3}{2}$.

Solution:

Given:

The sum of the first $p$ terms of an AP is $S_p = a$.

The sum of the first $q$ terms of the same AP is $S_q = b$.

The sum of the first $r$ terms of the same AP is $S_r = c$.

To Prove:

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

Proof:

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

The formula for the sum of the first $n$ terms of an AP is given by:

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

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

From equations (1), (2), and (3), we can express $\frac{a}{p}$, $\frac{b}{q}$, and $\frac{c}{r}$ as:

Consider the Left Hand Side (LHS) of the equation we need to prove:

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

Substitute the expressions for $\frac{a}{p}$, $\frac{b}{q}$, and $\frac{c}{r}$ from equations (4), (5), and (6) into the LHS:

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

Factor out $\frac{1}{2}$ from all terms:

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

Expand each term inside the parenthesis:

$[2A + (p-1)D](q-r) = 2A(q-r) + (p-1)D(q-r)$

$[2A + (q-1)D](r-p) = 2A(r-p) + (q-1)D(r-p)$

[2A + (r-1)D](p-q) = 2A(p-q) + (r-1)D(p-q)

Substitute these expansions back into the expression for LHS:

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

Group the terms containing $2A$ and the terms containing $D$:

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

Let's simplify the coefficients of $2A$ and $D$ separately.

Coefficient of $2A$:

$q-r + r-p + p-q = (q-q) + (-r+r) + (-p+p) = 0 + 0 + 0 = 0$

Coefficient of $D$:

Expand each product:

$(p-1)(q-r) = pq - pr - q + r$

$(q-1)(r-p) = qr - qp - r + p$

$(r-1)(p-q) = rp - rq - p + q$

Sum these three expanded terms:

$(pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q)$

$= pq - pr - q + r + qr - pq - r + p + rp - qr - p + q$

Group like terms:

$= (pq - pq) + (-pr + rp) + (-q + q) + (r - r) + (qr - qr) + (p - p)$

$= 0 + 0 + 0 + 0 + 0 + 0 = 0$

So, the coefficient of $D$ is also 0.

Substitute these simplified coefficients back into the expression for LHS:

$LHS = \frac{1}{2} \left( 2A(0) + D(0) \right)$

$LHS = \frac{1}{2} (0 + 0)$

$LHS = \frac{1}{2} (0)$

$LHS = 0$

The LHS is equal to 0, which is the Right Hand Side (RHS).

$LHS = RHS$

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

This completes the proof.

Solution:

Given:

Let the first AP be denoted by AP$_1$ and the second AP be denoted by AP$_2$.

Let the first term and common difference of AP$_1$ be $a_1$ and $d_1$ respectively.

Let the first term and common difference of AP$_2$ be $a_2$ and $d_2$ respectively.

The sum of the first $n$ terms of AP$_1$ is $S_n^{(1)}$, and the sum of the first $n$ terms of AP$_2$ is $S_n^{(2)}$.

The given ratio of their sums of $n$ terms is:

To Find:

The ratio of their 12th terms, i.e., $\frac{a_{12}^{(1)}}{a_{12}^{(2)}}$.

Solution:

The formula for the sum of the first $n$ terms of an AP with first term $a$ and common difference $d$ is $S_n = \frac{n}{2}[2a + (n-1)d]$.

Using this formula for the two APs, we have:

Now, consider the ratio of the sums:

$\frac{S_n^{(1)}}{S_n^{(2)}} = \frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]}$

The term $\frac{n}{2}$ cancels out, provided $n \neq 0$, which is true for the number of terms in a sum.

We are given that $\frac{S_n^{(1)}}{S_n^{(2)}} = \frac{3n+8}{7n+15}$. Equating this with the expression in (1):

Now, let's consider the ratio of the 12th terms of the two APs.

The formula for the $k$th term of an AP with first term $a$ and common difference $d$ is $a_k = a + (k-1)d$.

For the 12th term ($k=12$), the formula is $a_{12} = a + (12-1)d = a + 11d$.

So, the ratio of the 12th terms is:

Compare the expression in Equation (2) with the expression in Equation (3).

We have $\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2}$ in Equation (2) and $\frac{a_1 + 11d_1}{a_2 + 11d_2}$ in Equation (3).

If we can make the term $(n-1)$ in Equation (2) such that the numerator becomes proportional to $(a_1 + 11d_1)$ and the denominator becomes proportional to $(a_2 + 11d_2)$, we can find the ratio of the 12th terms.

Notice that $2a_1 + (n-1)d_1$ looks like $2 \times (a_1 + \frac{n-1}{2}d_1)$ and $a_1 + 11d_1$ is the 12th term.

If we set $\frac{n-1}{2} = 11$, then $n-1 = 2 \times 11 = 22$.

This means $n = 22 + 1 = 23$.

Let's substitute $n=23$ into Equation (2):

The factor of 2 cancels out from the numerator and the denominator on the left side:

From Equation (3), the left side of Equation (4) is the ratio of the 12th terms.

$\frac{a_{12}^{(1)}}{a_{12}^{(2)}} = \frac{77}{176}$

Now, simplify the fraction $\frac{77}{176}$. Both the numerator and the denominator are divisible by 11.

$\frac{77}{176} = \frac{11 \times 7}{11 \times 16}$

$\frac{\cancel{11}^{1} \times 7}{\cancel{11}_{1} \times 16} = \frac{7}{16}$

So, the ratio of the 12th terms is $\frac{7}{16}$.

The ratio of their 12th terms is 7 : 16.

Solution:

Given:

1. $a, b, c$ are in Arithmetic Progression (AP).

2. $b, c, d$ are in Geometric Progression (GP).

3. $c, d, e$ are in Harmonic Progression (HP).

To Show:

That $a, c, e$ are in Geometric Progression (GP).

Proof:

Let's translate the given conditions into algebraic equations based on the definitions of AP, GP, and HP.

Condition 1: $a, b, c$ are in AP.

This means that the difference between consecutive terms is constant. So, $b - a = c - b$.

Rearranging this gives: $2b = a + c$.

Condition 2: $b, c, d$ are in GP.

This means that the ratio of consecutive terms is constant. So, $\frac{c}{b} = \frac{d}{c}$.

Cross-multiplying gives: $c^2 = bd$.

Condition 3: $c, d, e$ are in HP.

This means that their reciprocals $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in AP.

So, the difference between consecutive reciprocal terms is constant: $\frac{1}{d} - \frac{1}{c} = \frac{1}{e} - \frac{1}{d}$.

Rearranging this gives: $\frac{1}{d} + \frac{1}{d} = \frac{1}{c} + \frac{1}{e}$.

$\frac{2}{d} = \frac{e+c}{ce}$

Taking the reciprocal of both sides: $\frac{d}{2} = \frac{ce}{e+c}$.

Multiplying by 2: $d = \frac{2ce}{e+c}$.

(Note: Since $c,d,e$ are in HP, none of them can be zero).

To show that $a, c, e$ are in GP, we need to prove that the square of the middle term $c$ is equal to the product of the other two terms, i.e., $c^2 = ae$.

Let's start with Equation (2) from the GP condition $b, c, d$:

From Equation (1), we can express $b$ in terms of $a$ and $c$:

From Equation (3), we have an expression for $d$ in terms of $c$ and $e$:

Now, substitute these expressions for $b$ and $d$ into the equation $c^2 = bd$:

Simplify the right side of the equation:

Multiply both sides of the equation by $(c+e)$:

Expand both sides of the equation:

$c^3 + c^2e = ace + c^2e$

Subtract $c^2e$ from both sides of the equation:

$c^3 + c^2e - c^2e = ace + c^2e - c^2e$

Since $c, d, e$ are in HP, $c$ cannot be zero. Therefore, we can divide both sides of the equation by $c$:

This is the condition for three numbers $a, c, e$ to be in Geometric Progression.

Therefore, $a, c, e$ are in GP.

This completes the proof.

Solution:

Given:

The two numbers are 3 and 24.

We need to insert 6 numbers between them such that the resulting sequence is an Arithmetic Progression (AP).

To Find:

The 6 numbers to be inserted between 3 and 24.

Solution:

Let the 6 numbers inserted between 3 and 24 be $A_1, A_2, A_3, A_4, A_5, A_6$.

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

This sequence is an Arithmetic Progression.

In this AP:

The first term is $a_1 = 3$.

The last term is $a_n = 24$.

The total number of terms in this AP is the first term + the 6 inserted means + the last term. So, the total number of terms is $n = 1 + 6 + 1 = 8$.

Thus, the last term, 24, is the 8th term of the AP, i.e., $a_8 = 24$.

The formula for the n-th term of an Arithmetic Progression is given by:

$a_n = a_1 + (n-1)d$

where $a_1$ is the first term and $d$ is the common difference.

Using the 8th term ($n=8$), we have:

Substitute the known values ($a_8=24$, $a_1=3$) into the equation:

Now, we solve Equation (1) for the common difference $d$:

Subtract 3 from both sides:

Divide both sides by 7:

Now that we have the first term ($a_1=3$) and the common difference ($d=3$), we can find the 6 inserted Arithmetic Means. The means are the terms $a_2, a_3, a_4, a_5, a_6, a_7$ of the AP.

First inserted number ($A_1 = a_2$):

$A_1 = a_1 + d = 3 + 3 = 6$

Second inserted number ($A_2 = a_3$):

$A_2 = a_1 + 2d = 3 + 2(3) = 3 + 6 = 9$

Third inserted number ($A_3 = a_4$):

$A_3 = a_1 + 3d = 3 + 3(3) = 3 + 9 = 12$

Fourth inserted number ($A_4 = a_5$):

$A_4 = a_1 + 4d = 3 + 4(3) = 3 + 12 = 15$

Fifth inserted number ($A_5 = a_6$):

$A_5 = a_1 + 5d = 3 + 5(3) = 3 + 15 = 18$

Sixth inserted number ($A_6 = a_7$):

$A_6 = a_1 + 6d = 3 + 6(3) = 3 + 18 = 21$

The 6 numbers to be inserted between 3 and 24 are 6, 9, 12, 15, 18, and 21.

The resulting AP sequence is 3, 6, 9, 12, 15, 18, 21, 24.

Solution:

Given:

The series is $5 + 55 + 555 + \dots$ up to $n$ terms.

To Find:

The sum of the first $n$ terms of the given series.

Solution:

Let the sum of the first $n$ terms of the series be $S_n$.

$S_n = 5 + 55 + 555 + \dots + a_n$

where $a_n$ is the $n$th term.

We can write each term by factoring out 5:

$S_n = 5(1) + 5(11) + 5(111) + \dots + 5(\underbrace{11\dots1}_{n \text{ times}})$

$S_n = 5(1 + 11 + 111 + \dots + \underbrace{11\dots1}_{n \text{ times}})$

Now, multiply and divide by 9 inside the parenthesis:

$S_n = 5 \times \frac{9}{9}(1 + 11 + 111 + \dots + \underbrace{11\dots1}_{n \text{ times}})$

$S_n = \frac{5}{9} (9 + 99 + 999 + \dots + \underbrace{99\dots9}_{n \text{ times}})$

Express each term in the parenthesis using powers of 10:

$9 = 10 - 1$

$99 = 100 - 1 = 10^2 - 1$

$999 = 1000 - 1 = 10^3 - 1$

... and so on.

The $n$th term is $\underbrace{99\dots9}_{n \text{ times}} = 10^n - 1$.

Substitute these expressions back into the sum:

$S_n = \frac{5}{9} [(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)]$

Rearrange the terms by grouping the powers of 10 and the -1 terms:

$S_n = \frac{5}{9} [(10 + 10^2 + 10^3 + \dots + 10^n) + (-1 - 1 - 1 - \dots - 1)]$

$S_n = \frac{5}{9} \left[ (10 + 10^2 + 10^3 + \dots + 10^n) - n \times 1 \right]$

$S_n = \frac{5}{9} \left[ (10 + 10^2 + 10^3 + \dots + 10^n) - n \right]$

The series $(10 + 10^2 + 10^3 + \dots + 10^n)$ is a Geometric Progression with:

First term $A = 10$.

Common ratio $R = \frac{10^2}{10} = 10$.

Number of terms = $n$.

The sum of the first $n$ terms of a GP with first term $A$ and common ratio $R$ is given by $S_n = A \frac{(R^n - 1)}{(R - 1)}$, where $R \neq 1$.

Using this formula for the GP $(10 + 10^2 + \dots + 10^n)$:

Sum of this GP $= 10 \frac{(10^n - 1)}{(10 - 1)} = 10 \frac{(10^n - 1)}{9}$

Substitute this sum back into the expression for $S_n$:

$S_n = \frac{5}{9} \left[ \frac{10(10^n - 1)}{9} - n \right]$

Simplify the expression by finding a common denominator inside the brackets:

$S_n = \frac{5}{9} \left[ \frac{10(10^n - 1)}{9} - \frac{9n}{9} \right]$

$S_n = \frac{5}{9} \left[ \frac{10(10^n - 1) - 9n}{9} \right]$

$S_n = \frac{5}{9} \left[ \frac{10^{n+1} - 10 - 9n}{9} \right]$

$S_n = \frac{5}{81} (10^{n+1} - 9n - 10)$

The sum of the first $n$ terms of the series $5 + 55 + 555 + \dots$ is $\frac{5}{81} (10^{n+1} - 9n - 10)$.

Solution:

Given:

The sum of the first three terms of a Geometric Progression (GP) is $\frac{13}{12}$.

The product of the first three terms of the GP is -1.

To Find:

The Geometric Progression (i.e., its first term and common ratio, or the first three terms).

Solution:

Let the first three terms of the Geometric Progression be $a$, $ar$, and $ar^2$, where $a$ is the first term and $r$ is the common ratio.

According to the given information, the sum of these terms is $\frac{13}{12}$:

And the product of these terms is -1:

Let's simplify the product equation (2):

$a \cdot a \cdot a \cdot r \cdot r^2 = -1$

$a^{1+1+1} r^{1+2} = -1$

$a^3 r^3 = -1$

$(ar)^3 = -1$

Taking the cube root of both sides to find the value of $ar$:

(Note: Since the product is -1, neither $a$ nor $r$ can be zero).

Now, substitute the value of $ar = -1$ into the sum equation (1):

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

Rearrange the terms to isolate $a-r$:

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

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

Now we have a system of two equations with two variables $a$ and $r$:

From (3): $ar = -1 \implies a = \frac{-1}{r}$ (since $r \neq 0$).

Substitute this expression for $a$ into equation (4):

Multiply the entire equation by $12r$ to clear the denominators (since $r \neq 0$):

$12r \left(\frac{-1}{r}\right) - 12r(r) = 12r \left(\frac{25}{12}\right)$

$-12 - 12r^2 = 25r$

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

We can solve this quadratic equation by factoring. We look for two numbers that multiply to $12 \times 12 = 144$ and add up to 25. These numbers are 9 and 16.

Rewrite the middle term $25r$ as $9r + 16r$:

$12r^2 + 9r + 16r + 12 = 0$

Group the terms and factor:

$(12r^2 + 9r) + (16r + 12) = 0$

$3r(4r + 3) + 4(4r + 3) = 0$

$(4r + 3)(3r + 4) = 0$

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

Case 1: $4r + 3 = 0 \implies 4r = -3 \implies r = -\frac{3}{4}$

Case 2: $3r + 4 = 0 \implies 3r = -4 \implies r = -\frac{4}{3}$

Now, find the corresponding first term $a$ for each value of $r$ using $a = \frac{-1}{r}$ from equation (3).

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

$a = \frac{-1}{-\frac{3}{4}} = -1 \times \left(-\frac{4}{3}\right) = \frac{4}{3}$

The first term is $a = \frac{4}{3}$ and the common ratio is $r = -\frac{3}{4}$.

The first three terms of this GP are:

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

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

$ar^2 = \frac{4}{3} \times (-\frac{3}{4})^2 = \frac{4}{3} \times \frac{9}{16} = \frac{\cancel{4}}{\cancel{3}} \times \frac{\cancel{9}^3}{\cancel{16}^4} = \frac{3}{4}$

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

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

$a = \frac{-1}{-\frac{4}{3}} = -1 \times \left(-\frac{3}{4}\right) = \frac{3}{4}$

The first term is $a = \frac{3}{4}$ and the common ratio is $r = -\frac{4}{3}$.

The first three terms of this GP are:

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

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

$ar^2 = \frac{3}{4} \times (-\frac{4}{3})^2 = \frac{3}{4} \times \frac{16}{9} = \frac{\cancel{3}}{\cancel{4}} \times \frac{\cancel{16}^4}{\cancel{9}^3} = \frac{4}{3}$

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

Both cases result in a GP whose first three terms satisfy the given conditions.

The Geometric Progression can be the one starting with $\frac{4}{3}$ and common ratio $-\frac{3}{4}$, or the one starting with $\frac{3}{4}$ and common ratio $-\frac{4}{3}$.

The first three terms of the GP are either $\frac{4}{3}, -1, \frac{3}{4}$ or $\frac{3}{4}, -1, \frac{4}{3}$.

Solution:

Given:

Two numbers are 1 and 256.

We need to insert 5 numbers between them such that the resulting sequence is a Geometric Progression (GP).

To Find:

The 5 numbers to be inserted between 1 and 256.

Solution:

Let the 5 numbers inserted between 1 and 256 be $G_1, G_2, G_3, G_4, G_5$.

The resulting sequence will be 1, $G_1, G_2, G_3, G_4, G_5$, 256.

This sequence is a Geometric Progression.

In this GP:

The first term is $a_1 = 1$.

The last term is $a_n = 256$.

The total number of terms in this GP is the first term + the 5 inserted means + the last term. So, the total number of terms is $n = 1 + 5 + 1 = 7$.

Thus, the last term, 256, is the 7th term of the GP, i.e., $a_7 = 256$.

The formula for the n-th term of a Geometric Progression is given by:

$a_n = a_1 r^{n-1}$

where $a_1$ is the first term and $r$ is the common ratio.

Using the 7th term ($n=7$), we have:

We need to find the values of $r$ that satisfy this equation. Taking the 6th root of 256:

$r = \pm \sqrt[6]{256}$

We know that $256 = 2^8$.

So, $r = \pm (2^8)^{1/6} = \pm 2^{8/6} = \pm 2^{4/3}$.

Thus, there are two possible values for the common ratio:

Case 1: $r = 2^{4/3}$

Case 2: $r = -2^{4/3}$

The 5 inserted numbers are the terms $a_2, a_3, a_4, a_5, a_6$ of the GP, which are given by $a_1 r, a_1 r^2, a_1 r^3, a_1 r^4, a_1 r^5$. Since $a_1 = 1$, the inserted numbers are $r, r^2, r^3, r^4, r^5$.

Case 1: Common ratio $r = 2^{4/3}$

$2^{4/3} = 2 \times 2^{1/3} = 2\sqrt[3]{2}$

$r^2 = (2^{4/3})^2 = 2^{8/3} = 2^2 \times 2^{2/3} = 4\sqrt[3]{4}$

$r^3 = (2^{4/3})^3 = 2^4 = 16$

$r^4 = (2^{4/3})^4 = 2^{16/3} = 2^5 \times 2^{1/3} = 32\sqrt[3]{2}$

$r^5 = (2^{4/3})^5 = 2^{20/3} = 2^6 \times 2^{2/3} = 64\sqrt[3]{4}$

The 5 inserted numbers are $2\sqrt[3]{2}, 4\sqrt[3]{4}, 16, 32\sqrt[3]{2}, 64\sqrt[3]{4}$.

The resulting GP is $1, 2\sqrt[3]{2}, 4\sqrt[3]{4}, 16, 32\sqrt[3]{2}, 64\sqrt[3]{4}, 256$.

Case 2: Common ratio $r = -2^{4/3}$

$r = -2^{4/3} = -2\sqrt[3]{2}$

$r^2 = (-2^{4/3})^2 = 2^{8/3} = 4\sqrt[3]{4}$

$r^3 = (-2^{4/3})^3 = -2^4 = -16$

$r^4 = (-2^{4/3})^4 = 2^{16/3} = 32\sqrt[3]{2}$

$r^5 = (-2^{4/3})^5 = -2^{20/3} = -64\sqrt[3]{4}$

The 5 inserted numbers are $-2\sqrt[3]{2}, 4\sqrt[3]{4}, -16, 32\sqrt[3]{2}, -64\sqrt[3]{4}$.

The resulting GP is $1, -2\sqrt[3]{2}, 4\sqrt[3]{4}, -16, 32\sqrt[3]{2}, -64\sqrt[3]{4}, 256$. (Note the alternating signs).

The 5 numbers to be inserted between 1 and 256 are either $2\sqrt[3]{2}, 4\sqrt[3]{4}, 16, 32\sqrt[3]{2}, 64\sqrt[3]{4}$ or $-2\sqrt[3]{2}, 4\sqrt[3]{4}, -16, 32\sqrt[3]{2}, -64\sqrt[3]{4}$.

Solution:

Given:

Let the two positive numbers be $x$ and $y$.

The Arithmetic Mean (AM) of $x$ and $y$ is $A$.

The Geometric Mean (GM) of $x$ and $y$ is $G$.

To Prove:

The two positive numbers are $A + \sqrt{A^2 - G^2}$ and $A - \sqrt{A^2 - G^2}$.

Proof:

By the definitions of Arithmetic Mean and Geometric Mean for two positive numbers $x$ and $y$:

We are given that $AM = A$ and $GM = G$. So, we have the following equations:

From Equation (1), we can express the sum of the two numbers:

From Equation (2), we can express the product of the two numbers by squaring both sides:

We now have the sum and the product of the two numbers $x$ and $y$. Consider a quadratic equation whose roots are $x$ and $y$. Such a quadratic equation is given by:

Substitute the sum $(x+y) = 2A$ from Equation (3) and the product $xy = G^2$ from Equation (4) into the quadratic equation:

This is a quadratic equation in the variable $t$. The roots of this equation are the two numbers $x$ and $y$ we are looking for.

We can solve this quadratic equation using the quadratic formula for roots of $at^2 + bt + c = 0$, which is $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

In Equation (5), we have $a=1$, $b=-2A$, and $c=G^2$.

Substitute these values into the quadratic formula:

Factor out 4 from the term under the square root:

Factor out 2 from the numerator:

Cancel the common factor of 2:

The two roots of this quadratic equation are the two numbers $x$ and $y$. Therefore, the two numbers are $A + \sqrt{A^2 - G^2}$ and $A - \sqrt{A^2 - G^2}$.

For the numbers to be real, the discriminant must be non-negative: $4A^2 - 4G^2 \ge 0$, which simplifies to $A^2 \ge G^2$. Since $A$ and $G$ are the AM and GM of positive numbers, $A > 0$ and $G > 0$. The AM-GM inequality states that for positive numbers, $A \ge G$. Squaring both sides of $A \ge G$ gives $A^2 \ge G^2$. Thus, the term under the square root is always non-negative when $A$ and $G$ are the AM and GM of positive numbers.

The two numbers are indeed $A + \sqrt{A^2 - G^2}$ and $A - \sqrt{A^2 - G^2}$.

This completes the proof.

Solution:

Given:

The $n$th term of the series is given by $a_n = n(n+1)(n+4)$.

To Find:

The sum of the series up to $n$ terms, denoted by $S_n = \sum\limits_{k=1}^{n} a_k$.

Solution:

The $k$th term of the series is $a_k = k(k+1)(k+4)$.

First, we expand the expression for the $k$th term:

$a_k = (k^2 + k)(k+4)$

$a_k = k^2(k+4) + k(k+4)$

$a_k = k^3 + 4k^2 + k^2 + 4k$

$a_k = k^3 + 5k^2 + 4k$

The sum of the first $n$ terms is $S_n = \sum\limits_{k=1}^{n} a_k$.

$S_n = \sum\limits_{k=1}^{n} (k^3 + 5k^2 + 4k)$

Using the properties of summation, we can write the sum as:

$S_n = \sum\limits_{k=1}^{n} k^3 + \sum\limits_{k=1}^{n} 5k^2 + \sum\limits_{k=1}^{n} 4k$

$S_n = \sum\limits_{k=1}^{n} k^3 + 5 \sum\limits_{k=1}^{n} k^2 + 4 \sum\limits_{k=1}^{n} k$

Now, we use the standard formulas for the sums of powers of the first $n$ natural numbers:

• $\sum\limits_{k=1}^{n} k = \frac{n(n+1)}{2}$
• $\sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
• $\sum\limits_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^2(n+1)^2}{4}$

Substitute these formulas into the expression for $S_n$:

$S_n = \frac{n^2(n+1)^2}{4} + 5 \left(\frac{n(n+1)(2n+1)}{6}\right) + 4 \left(\frac{n(n+1)}{2}\right)$

$S_n = \frac{n^2(n+1)^2}{4} + \frac{5n(n+1)(2n+1)}{6} + \frac{4n(n+1)}{2}$

$S_n = \frac{n^2(n+1)^2}{4} + \frac{5n(n+1)(2n+1)}{6} + 2n(n+1)$

To combine these terms, find a common denominator, which is 12.

$S_n = \frac{3 \cdot n^2(n+1)^2}{12} + \frac{2 \cdot 5n(n+1)(2n+1)}{12} + \frac{12 \cdot 2n(n+1)}{12}$

$S_n = \frac{3n^2(n+1)^2 + 10n(n+1)(2n+1) + 24n(n+1)}{12}$

Factor out the common term $n(n+1)$ from the numerator:

$S_n = \frac{n(n+1) [3n(n+1) + 10(2n+1) + 24]}{12}$

Expand the terms inside the square brackets:

$3n(n+1) = 3n^2 + 3n$

$10(2n+1) = 20n + 10$

Substitute these back into the expression inside the brackets:

$3n^2 + 3n + 20n + 10 + 24$

$= 3n^2 + (3n + 20n) + (10 + 24)$

$= 3n^2 + 23n + 34$

Substitute this quadratic expression back into the formula for $S_n$:

$S_n = \frac{n(n+1) (3n^2 + 23n + 34)}{12}$

We can factor the quadratic expression $3n^2 + 23n + 34$. We look for two numbers that multiply to $3 \times 34 = 102$ and add up to 23. These numbers are 6 and 17.

$3n^2 + 23n + 34 = 3n^2 + 6n + 17n + 34$

$= 3n(n+2) + 17(n+2)$

$= (n+2)(3n+17)$

Substitute the factored quadratic expression back into the formula for $S_n$:

$S_n = \frac{n(n+1)(n+2)(3n+17)}{12}$

The sum of the series whose $n$th term is $a_n = n(n+1)(n+4)$ is $\frac{n(n+1)(n+2)(3n+17)}{12}$.

Solution:

Given:

Savings in the first month = $\textsf{₹} 200$.

Savings in the second month = $\textsf{₹} 250$.

Savings in the third month = $\textsf{₹} 300$.

This pattern of savings forms an Arithmetic Progression (AP).

Total savings accumulated = $\textsf{₹} 39000$.

To Find:

The number of months ($n$) in which his total savings will be $\textsf{₹} 39000$.

Solution:

The monthly savings form an Arithmetic Progression with:

First term, $a = 200$.

Common difference, $d = 250 - 200 = 50$.

Let $S_n$ be the total savings after $n$ months. We are given $S_n = 39000$.

The formula for the sum of the first $n$ terms of an Arithmetic Progression is given by:

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

Substitute the given values of $S_n = 39000$, $a = 200$, and $d = 50$ into the formula:

Now, we solve this equation for $n$.

$39000 = \frac{n}{2}[400 + 50n - 50]$

$39000 = \frac{n}{2}[350 + 50n]$

Multiply both sides by 2:

$2 \times 39000 = n(350 + 50n)$

$78000 = 350n + 50n^2$

Rearrange the equation into a standard quadratic form $An^2 + Bn + C = 0$:

$50n^2 + 350n - 78000 = 0$

Divide the entire equation by 50 to simplify:

$\frac{50n^2}{50} + \frac{350n}{50} - \frac{78000}{50} = 0$

This is a quadratic equation in the variable $n$. We can use the quadratic formula to find the values of $n$. The quadratic formula for an equation $an^2 + bn + c = 0$ is $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

In our equation (2), $a=1$, $b=7$, and $c=-1560$.

Substitute these values into the quadratic formula:

The number of months must be a positive integer. We need to determine if the positive root of this equation is an integer.

The discriminant is $6289$. Let's check if 6289 is a perfect square.

We know $79^2 = 6241$ and $80^2 = 6400$. Since $6241 < 6289 < 6400$, the square root of 6289 is between 79 and 80. Therefore, $\sqrt{6289}$ is not an integer.

The two possible values for $n$ are:

$n_1 = \frac{-7 + \sqrt{6289}}{2}$

$n_2 = \frac{-7 - \sqrt{6289}}{2}$

Since the number of months must be positive, we consider the positive root:

As $\sqrt{6289}$ is not an integer, the value of $n$ is also not an integer.

This means that based on the given savings pattern, the total savings of $\textsf{₹} 39000$ is not achieved exactly at the end of a whole number of months.

Based on the provided numbers, the number of months required for the total savings to be exactly $\textsf{₹} 39000$ is $\frac{-7 + \sqrt{6289}}{2}$, which is not a whole number.

(Note: If the question implies finding the first whole number of months where the savings meet or exceed $\textsf{₹} 39000$, we would find $S_{36} = \textsf{₹} 38700$ and $S_{37} = \textsf{₹} 40700$. In that case, the savings first exceed $\textsf{₹} 39000$ after 37 months. However, the question asks for the number of months when the savings "will be $\textsf{₹} 39000$", which leads to the non-integer solution derived above.)

Solution:

Given:

Let the Geometric Progression be $a, ar, ar^2, \dots, ar^{n-1}$.

The first term of the GP is $a_1 = a$.

The $n$th term of the GP 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 Geometric Progression be $r$.

The formula for the $k$th term of a GP with first term $a_1$ and common ratio $r$ is $a_k = a_1 r^{k-1}$.

Given that the first term is $a$, we have $a_1 = a$.

The $n$th term is given as $b$. Using the formula for the $n$th term ($k=n$), we have:

Substituting the given values $a_1 = a$ and $a_n = b$:

The product of the first $n$ terms of the GP is $P$.

$P = a_1 \times a_2 \times a_3 \times \dots \times a_n$

$P = a \times (ar) \times (ar^2) \times \dots \times (ar^{n-1})$

Group the terms with $a$ and the terms with $r$:

$P = (a \times a \times a \times \dots \times a \text{ ($n$ times)}) \times (r^0 \times r^1 \times r^2 \times \dots \times r^{n-1})$

$P = a^n \times r^{(0+1+2+\dots+(n-1))}$

The sum in the exponent of $r$ is the sum of the first $(n-1)$ non-negative integers. This sum is given by the formula for the sum of an arithmetic series: $S_{m} = \frac{m(m+1)}{2}$ where $m=n-1$.

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

Substitute this sum back into the expression for $P$:

Now, we want to relate this expression for $P$ to $a$ and $b$. From Equation (1), we have $b = ar^{n-1}$. This implies $r^{n-1} = \frac{b}{a}$.

Let's rewrite the expression for $P$ (Equation 2) by manipulating the exponent of $r$:

$P = a^n r^{n \times \frac{(n-1)}{2}}$

$P = a^n (r^{n-1})^{\frac{n}{2}}$

Substitute $r^{n-1} = \frac{b}{a}$ from Equation (1) into this expression:

$P = a^n \frac{b^{n/2}}{a^{n/2}}$

$P = a^{n - \frac{n}{2}} b^{\frac{n}{2}}$

$P = a^{\frac{n}{2}} b^{\frac{n}{2}}$

Finally, we need to prove $P^2 = (ab)^n$. Square both sides of Equation (3):

$P^2 = (ab)^{\frac{n}{2} \times 2}$

Thus, we have shown that the square of the product of the first $n$ terms is equal to the $n$th power of the product of the first and $n$th terms.

This completes the proof.

Solution:

Given:

The series is $\frac{1^2}{1} + \frac{1^2+2^2}{1+2} + \frac{1^2+2^2+3^2}{1+2+3} + \dots$

To Find:

The sum of the first $n$ terms of the series.

Solution:

Let the $k$th term of the series be denoted by $a_k$.

The numerator of the $k$th term is the sum of the squares of the first $k$ natural numbers: $1^2 + 2^2 + \dots + k^2 = \sum\limits_{i=1}^{k} i^2$.

The formula for the sum of the squares of the first $k$ natural numbers is:

$\sum\limits_{i=1}^{k} i^2 = \frac{k(k+1)(2k+1)}{6}$

The denominator of the $k$th term is the sum of the first $k$ natural numbers: $1 + 2 + \dots + k = \sum\limits_{i=1}^{k} i$.

The formula for the sum of the first $k$ natural numbers is:

$\sum\limits_{i=1}^{k} i = \frac{k(k+1)}{2}$

Now, we find the expression for the $k$th term, $a_k$, by dividing the numerator sum by the denominator sum:

$a_k = \frac{\text{Sum of squares of first } k \text{ natural numbers}}{\text{Sum of first } k \text{ natural numbers}} = \frac{\sum\limits_{i=1}^{k} i^2}{\sum\limits_{i=1}^{k} i}$

Substitute the formulas for the sums:

$a_k = \frac{\frac{k(k+1)(2k+1)}{6}}{\frac{k(k+1)}{2}}$

To simplify this complex fraction, we multiply the numerator by the reciprocal of the denominator:

$a_k = \frac{k(k+1)(2k+1)}{6} \times \frac{2}{k(k+1)}$

Assuming $k \ge 1$, $k(k+1)$ is non-zero, so we can cancel out the common factor $k(k+1)$:

$a_k = \frac{\cancel{k(k+1)}(2k+1)}{6} \times \frac{2}{\cancel{k(k+1)}}$

$a_k = \frac{2(2k+1)}{6}$

Simplify the fraction by dividing the numerator and denominator by 2:

$a_k = \frac{2k+1}{3}$

We need to find the sum of the first $n$ terms of this series, which is $S_n = \sum\limits_{k=1}^{n} a_k$.

Substitute the expression for $a_k$:

$S_n = \sum\limits_{k=1}^{n} \frac{2k+1}{3}$

Using the property of summation $\sum\limits_{k=1}^{n} (c_1 f(k) + c_2 g(k)) = c_1 \sum\limits_{k=1}^{n} f(k) + c_2 \sum\limits_{k=1}^{n} g(k)$, we can split the sum:

$S_n = \sum\limits_{k=1}^{n} \left(\frac{2k}{3} + \frac{1}{3}\right)$

$S_n = \sum\limits_{k=1}^{n} \frac{2k}{3} + \sum\limits_{k=1}^{n} \frac{1}{3}$

$S_n = \frac{2}{3} \sum\limits_{k=1}^{n} k + \frac{1}{3} \sum\limits_{k=1}^{n} 1$

Now, we use the standard formulas for the sum of the first $n$ natural numbers ($\sum\limits_{k=1}^{n} k$) and the sum of a constant $n$ times ($\sum\limits_{k=1}^{n} c = nc$):

$\sum\limits_{k=1}^{n} k = \frac{n(n+1)}{2}$

$\sum\limits_{k=1}^{n} 1 = n$

Substitute these formulas back into the expression for $S_n$:

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

$S_n = \frac{\cancel{2}}{3} \cdot \frac{n(n+1)}{\cancel{2}} + \frac{n}{3}$

$S_n = \frac{n(n+1)}{3} + \frac{n}{3}$

Since both terms have a common denominator of 3, we can add the numerators:

$S_n = \frac{n(n+1) + n}{3}$

Expand the term $n(n+1)$ in the numerator:

$S_n = \frac{n^2 + n + n}{3}$

$S_n = \frac{n^2 + 2n}{3}$

We can factor out $n$ from the terms in the numerator:

$S_n = \frac{n(n+2)}{3}$

The sum of the first $n$ terms of the given series is $\frac{n(n+2)}{3}$.