If in an A.P., Sₙ = qn² and Sm = qm², where Sr denotes the sum of r terms of the A.P., then Sq equals (A) (q³)

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S_n = qn^2

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Concept understanding — Geometric Progression

Geometric Progression: The Idea of Repeated Multiplication

Imagine you're folding a piece of paper in half. Start with thickness 1 unit. After one fold, thickness becomes 2. After two folds, thickness becomes 4. After three folds, thickness becomes 8. The sequence of thicknesses is:

1, 2, 4, 8, 16, ...

Notice the pattern: each term is obtained by multiplying the previous term by the same number (here, 2). That's the core intuition behind a geometric progression — you keep multiplying by a fixed number, step after step.

This is different from an arithmetic progression, where you keep adding a fixed number. Here, the growth is multiplicative, not additive. That's why geometric progressions grow (or shrink) much faster.

Precise Definition

A Geometric Progression (GP) is a sequence of numbers where the ratio of any term to its preceding term is constant. This constant is called the common ratio, denoted by $r$ .

If the first term is $a$ , then the sequence looks like:

a, \ ar, \ ar^2, \ ar^3, \ ar^4, \ \dots

Note

The common ratio $r$ can be any real number — positive, negative, or even a fraction. If $r$ is negative, the terms alternate in sign. If $0 < r < 1$ , the terms get smaller and smaller.

The $n$ -th Term

To find any term directly without listing all previous ones, use the formula:

T_n = a \cdot r^{\,n-1}

where $T_n$ is the $n$ -th term, $a$ is the first term, $r$ is the common ratio, and $n$ is the term number (starting from 1).

Example: For the paper-folding sequence, $a = 1$ , $r = 2$ . The 5th term is $1 \cdot 2^{5-1} = 2^4 = 16$ , which matches our list.

Sum of $n$ Terms

There are two cases, depending on whether $r = 1$ or not.

n

When $r = 1$ , every term is just $a$ , so the sum is simply $n \times a$ .

Why the formula works (intuition):

Let $S = a + ar + ar^2 + \dots + ar^{n-1}$ . Multiply both sides by $r$ : $rS = ar + ar^2 + \dots + ar^n$ . Subtract the first from the second: $rS - S = ar^n - a$ , so $S(r-1) = a(r^n - 1)$ , giving the formula above.

Sum of an Infinite GP

If the common ratio $r$ lies strictly between $-1$ and $1$ (i.e., $|r| < 1$ ), the terms get smaller and smaller, and the sum of all terms approaches a finite value:

S_\infty = \frac{a}{1 - r}, \quad \text{for } |r| < 1

Watch out

If $|r| \geq 1$ , the infinite sum does not exist (it diverges to infinity or oscillates without settling). Never apply the infinite sum formula when $|r| \geq 1$ .

Example: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ has $a=1$ , $r=\frac12$ , so $S_\infty = \frac{1}{1 - 1/2} = 2$ . This matches the intuition that repeatedly halving a unit length eventually fills exactly 2 units.

Quick Reference Table

Property	Formula	Condition
Common ratio	$r = \dfrac{T_{n+1}}{T_n}$	Always
$n$ -th term	$T_n = a r^{n-1}$	Always

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The sum of $r$ terms of an AP, $S_r$ , is generally a quadratic in $r$ . By comparing the given $S_n = qn^2$ with the general form, we deduce that $S_r = qr^2$ for any $r$ . Substituting $r=q$ then gives $S_q = \boxed{q^3}$ .

The sum of the first $r$ terms of an arithmetic progression (AP) is a fundamental concept. Understanding its general structure is key to solving this problem efficiently.

Concept and Intuition

The sum of the first $r$ terms of an AP with first term $a$ and common difference $d$ is given by the formula:

S_r = \frac{r}{2}[2a + (r-1)d]

Let's expand this expression to see its structure:

S_r = \frac{r}{2}(2a + rd - d)

S_r = ar + \frac{dr^2}{2} - \frac{dr}{2}

Rearranging the terms by powers of $r$ :

S_r = \left(\frac{d}{2}\right)r^2 + \left(a - \frac{d}{2}\right)r

This shows that $S_r$ is always a quadratic expression in $r$ of the form $Ar^2 + Br$ , where $A = \frac{d}{2}$ and $B = a - \frac{d}{2}$ .

The problem states that $S_n = qn^2$ . This means that the sum of $r$ terms, $S_r$ , follows the pattern $qr^2$ . Comparing this specific form with the general quadratic form $Ar^2 + Br$ , we can deduce the values of $A$ and $B$ . In $S_r = qr^2$ , the coefficient of $r^2$ is $q$ , and the coefficient of $r$ is $0$ . This insight allows us to determine the first term ( $a$ ) and common difference ( $d$ ) of the AP in terms of $q$ . Once we know the general formula for $S_r$ , finding $S_q$ is a straightforward substitution.

Step-by-step Derivation

Identify the general form of $S_r$ for an AP: As derived above, the sum of the first $r$ terms of an AP with first term $a$ and common difference $d$ is:

S_r = \left(\frac{d}{2}\right)r^2 + \left(a - \frac{d}{2}\right)r

Compare with the given information: The problem states that $S_n = qn^2$ . This implies that the sum of $r$ terms of this specific AP is given by the formula $S_r = qr^2$ . We now equate the coefficients of the general form with the given form:

qr^2 = \left(\frac{d}{2}\right)r^2 + \left(a - \frac{d}{2}\right)r

For these two expressions to be identical for all values of $r$, the coefficients of corresponding powers of $r$ must be equal. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AHSEC Higher Secondary (HS) 1st Year Examination 2026Set ANNUAL1 markMCQ
$4$
›Reveal solutionSolution

Product of first $5$ GP terms $= (t_3)^5 = 4^5$ .

Let the GP be $a, ar, ar^2, ar^3, ar^4$ .

Product $= a^5 r^{0+1+2+3+4} = a^5 r^{10} = (ar^2)^5$ .

…

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AHSEC Higher Secondary (HS) 1st Year Examination 2026Set ANNUAL1 markMCQ
$1$
›Reveal solutionSolution

$1, G_1, G_2, 64$ in GP: $r^3 = 64 \Rightarrow r = 4$ , so $G_1=4,\ G_2=16$ .

Two geometric means $G_1, G_2$ between $1$ and $64$ form the GP $1, G_1, G_2, 64$ .

…

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AHSEC Higher Secondary (HS) 1st Year Examination 2026Set ANNUAL1 markMCQ
$9^{\frac{1}{3}} \cdot 9^{\frac{1}{9}} \cdot 9^{\frac{1}{27}} \cdots \infty$
›Reveal solutionSolution

Add the exponents ( $\frac12$ ) then evaluate: $9^{1/2}=3$ .

Using $9^{a}\cdot 9^{b} \cdots = 9^{a+b+\cdots}$ , add the exponents:

$S = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots$ is an infinite GP with first term $\frac13$ and ratio $\frac13$ .

…

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AHSEC Higher Secondary (HS) 1st Year Examination 2025Set ANNUAL1 markMCQ
$2, 8, 32, 128, \ldots$
(a) $8$
(b) $9$
(c) $7$
(d) None
›Reveal solutionSolution

The $9$ th term equals $131072$ .

For the GP, $a=2$ and $r=\frac{8}{2}=4$ . The $n$ th term is

$T_n=a\,r^{\,n-1}=2\cdot4^{\,n-1}.$

Set $T_n=131072$ : …

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AHSEC Higher Secondary (HS) 1st Year Examination 2023Set ANNUAL1 mark
$x$
›Reveal solutionSolution

In a GP, the square of the middle term equals the product of its neighbours.

For three numbers $a, x, b$ to be in GP, the middle term squared equals the product of the outer terms: $x^2 = a \cdot b$ .

Here $a = -\dfrac{2}{7}$ , $b = -\dfrac{7}{2}$ .

…

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AHSEC Higher Secondary (HS) 1st Year Examination 2022Set ANNUAL1 mark
$3, \frac{3}{2}, \frac{3}{4}, \ldots$
›Reveal solutionSolution

In a GP, the common ratio is any term divided by the term immediately before it.

The series is $3,\ \dfrac{3}{2},\ \dfrac{3}{4},\ldots$

$r = \frac{a_2}{a_1} = \frac{3/2}{3} = \frac{1}{2}$

…

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Q.If in an A.P., Sn=qn2S_n = qn^2Sn​=qn2 and Sm=qm2S_m = qm^2Sm​=qm2, where SrS_rSr​ denotes the sum of rrr terms of the A.P., then SqS_qSq​ equals (A) q32\dfrac{q^3}{2}2q3​ (B) mnqmnqmnq (C) q3q^3q3 (D) (m+n)q2(m + n)q^2(m+n)q2

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Q.If in an A.P., $S_n = qn^2$ and $S_m = qm^2$ , where $S_r$ denotes the sum of $r$ terms of the A.P., then $S_q$ equals
(A) $\dfrac{q^3}{2}$
(B) $mnq$
(C) $q^3$
(D) $(m + n)q^2$

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