Derive sum formula for geometric progression. When |r| < 1, S∞ = a/(1-r).

Sum of n Terms of GP — Sₙ = a(rⁿ - 1)/(r - 1)

Question

Derive the formula for the sum of the first $n$ terms of a Geometric Progression (GP). Also state the formula for infinite GP when $|r| < 1$ .

Given a GP with first term $a$ and common ratio $r$ , find $S_n = a + ar + ar^2 + \cdots + ar^{n-1}$ .

Solution — Step by Step

Let $S_n$ be the sum of the first $n$ terms:

S_n = a + ar + ar^2 + ar^3 + \cdots + ar^{n-1}

This is just every term written out — nothing hidden yet.

Here’s the trick that makes the whole derivation work. Multiply the entire equation by $r$ :

rS_n = ar + ar^2 + ar^3 + \cdots + ar^{n-1} + ar^n

Notice how $rS_n$ looks almost identical to $S_n$ , just shifted by one term.

Now subtract the second equation from the first:

S_n - rS_n = a - ar^n

Almost every middle term cancels — only the first term of $S_n$ and the last term of $rS_n$ survive. This telescoping is the heart of the derivation.

S_n(1 - r) = a(1 - r^n)

Divide both sides by $(1 - r)$ , valid when $r \neq 1$ :

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

You’ll also see this written as $\dfrac{a(r^n - 1)}{r - 1}$ — both are the same, just multiplied by $\frac{-1}{-1}$ .

When $r = 1$ , every term equals $a$ , so:

S_n = na

This isn’t a GP in the usual sense (no ratio variation), but NCERT expects you to state it separately.

Why This Works

The subtraction trick works because a GP has a rigid multiplicative structure — each term is exactly $r$ times the previous one. When you multiply $S_n$ by $r$ , you’re essentially shifting the entire sequence forward by one position.

The “telescoping” that happens on subtraction is the same idea used in telescoping series across calculus too. You’ll see this pattern again when summing $\sum \frac{1}{k(k+1)}$ — recognise it early and it saves a lot of time.

For the infinite GP when $|r| < 1$ : as $n \to \infty$ , the term $r^n \to 0$ . So:

S_\infty = \frac{a}{1 - r} \quad \text{(valid only when } |r| < 1\text{)}

This shows up directly in JEE Main questions about recurring decimals like $0.\overline{3} = 0.3 + 0.03 + \cdots$

Alternative Method

Using the pattern directly for infinite GP:

Write $S = a + ar + ar^2 + \cdots$

Multiply by $r$ : $rS = ar + ar^2 + ar^3 + \cdots$

Subtract: $S - rS = a$ , so $S(1-r) = a$ , giving $S = \dfrac{a}{1-r}$ .

This is quicker for infinite GP problems and avoids the $r^n$ term entirely. In JEE, when you see “sum to infinity” in the question, jump straight to this version.

To decide which form of the finite formula to use: if $r > 1$ , use $\dfrac{a(r^n - 1)}{r-1}$ (positive numerator and denominator). If $r < 1$ , use $\dfrac{a(1 - r^n)}{1-r}$ . Same answer, but avoids sign errors under exam pressure.

Common Mistake

Applying the infinite GP formula when $|r| \geq 1$ . Students write $S_\infty = \dfrac{a}{1-r}$ even when $r = 2$ or $r = -2$ . The infinite sum only converges when $|r| < 1$ — otherwise the terms keep growing and there’s no finite sum. If $r = 1$ , the sum is just $na$ which blows up. If $r = -1$ , the sum oscillates and never settles. Always check $|r| < 1$ before using $S_\infty$ .

Case	Formula
Finite GP, any $r \neq 1$	$S_n = \dfrac{a(1 - r^n)}{1 - r}$
Finite GP, $r = 1$	$S_n = na$
Infinite GP, $\\|r\\| < 1$	$S_\infty = \dfrac{a}{1 - r}$

Here $a$ = first term, $r$ = common ratio, $n$ = number of terms.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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