If the sum of $n$ terms of an A.P. is given by $S_n = 3n + 2n^2$, then the common difference of the A.P. is (A) $3$ (B) $2$ (C) $6$ (D) $4$

If the sum of n terms of an A.P. is given by Sₙ = 3n + 2n², then the common difference of the A.P. is (A) 3 (B

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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 formula $S_n = 3n + 2n^2$ is quadratic in $n$ , which is characteristic of an AP. By comparing with the standard form $S_n = \frac{n}{2}[2a + (n-1)d]$ , or by finding the first term and second term directly, the common difference comes out to be $4$ .

The key idea here is that the sum of $n$ terms of an arithmetic progression is always a quadratic function of $n$ with no constant term. Specifically, if the first term is $a$ and the common difference is $d$ , then

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

So the coefficient of $n^2$ is $\frac{d}{2}$ , and the coefficient of $n$ is $a - \frac{d}{2}$ . Given $S_n = 2n^2 + 3n$ , we can match coefficients directly.

But let’s also do it step-by-step from first principles — that builds stronger intuition.

Find the first term $a$ . The first term $a$ is simply $S_1$ , because the sum of 1 term is that term itself.

a = S_1 = 3(1) + 2(1)^2 = 3 + 2 = 5.

Find the second term $a_2$ . The sum of the first two terms is $S_2$ .

S_2 = 3(2) + 2(2)^2 = 6 + 8 = 14.

Since $S_2 = a + a_2$ , we have

5 + a_2 = 14 \quad\Rightarrow\quad a_2 = 9.

The common difference $d$ is the difference between consecutive terms. …

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Q.If the sum of nnn terms of an A.P. is given by Sn=3n+2n2S_n = 3n + 2n^2Sn​=3n+2n2, then the common difference of the A.P. is (A) 333 (B) 222 (C) 666 (D) 444

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Q.If the sum of $n$ terms of an A.P. is given by $S_n = 3n + 2n^2$ , then the common difference of the A.P. is
(A) $3$
(B) $2$
(C) $6$
(D) $4$

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