Sum of GP: 2 + 6 + 18 + ... up to 6 Terms — Geometric Series

easy CBSE JEE-MAIN NCERT Class 11 Chapter 9 3 min read

Question

Find the sum of the geometric progression 2 + 6 + 18 + … up to 6 terms.


Solution — Step by Step

The first term is a = 2. The common ratio is r = 6/2 = 3 (each term is 3 times the previous). We need n = 6 terms.

Quick check: 2 → 6 → 18 → 54 — yes, ratio is consistently 3.

Since r ≠ 1, we use:

Sn=a(rn1)r1S_n = \frac{a(r^n - 1)}{r - 1}

We use this form (not the other one with 1 - rⁿ) because r > 1, which keeps everything positive and avoids sign confusion.

S6=2(361)31S_6 = \frac{2(3^6 - 1)}{3 - 1}

First, calculate 3⁶. We build it up: 3² = 9, 3³ = 27, 3⁶ = 27² = 729.

S6=2(7291)2=2×7282S_6 = \frac{2(729 - 1)}{2} = \frac{2 \times 728}{2}
S6=14562=728S_6 = \frac{1456}{2} = 728

The 2 in the numerator and denominator cancel cleanly — this happens often when a = 2 and r = 3.

The sum of the first 6 terms is 728.


Why This Works

A geometric series grows by a fixed multiplier each step, so you can’t just add terms with an arithmetic shortcut. The formula Sn = a(rⁿ - 1)/(r - 1) comes from a neat algebraic trick: write S, then write rS (the series shifted by one term), and subtract. Almost everything cancels, leaving just the first and last terms.

This is why the formula has rⁿ in it — the subtraction isolates the “gap” between the scaled series and the original. Understanding this derivation means you’ll never confuse the two forms of the formula under exam pressure.

For this question, 3⁶ = 729 is the number to get right. In JEE-pattern MCQs, wrong powers of 3 are the most common trap answer.


Alternative Method

You can verify by adding terms directly — useful as a sanity check.

TermValueRunning Sum
1st22
2nd68
3rd1826
4th5480
5th162242
6th486728

This matches our formula answer. Direct addition is fine for small n, but for n = 20 or when the question asks for a general n, you need the formula.

When r = 3, memorise the powers: 3, 9, 27, 81, 243, 729, 2187. These appear constantly in NCERT exercises and board papers. 3⁶ = 729 is the most-tested one.


Common Mistake

Students write S_n = a(1 - rⁿ)/(1 - r) when r > 1, which gives the same numerical answer but with a negative numerator and negative denominator. That’s fine mathematically, but panicking mid-calculation and dropping a sign gives a negative sum — clearly wrong for a series of positive terms. Fix: when r > 1, always use a(rⁿ - 1)/(r - 1). When r < 1, use a(1 - rⁿ)/(1 - r). Match the form to the sign of (r - 1) so you’re never dividing by a negative.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next