Geometric Progression • Topic 2 of 3

Sum of GP

The sum of the first n terms of a GP (with r ≠ 1) is S_n = a(r^n − 1)/(r − 1). When r > 1 this form keeps the numbers positive; when r < 1 the equivalent S_n = a(1 − r^n)/(1 − r) is tidier. If r = 1 every term equals a, so the sum is simply n·a — a special case CAT sometimes slips in to catch the unwary. The fastest approach in the exam is to compute r^n first using powers you already know (2^10 = 1024, 3^4 = 81), then plug in. A useful manipulation: S_n(r − 1) = a(r^n − 1), which lets you solve for an unknown a, r or n without dividing early and accumulating fractions. Recognise the sum of a GP inside compound-interest annuities and inside questions that ask for 1 + 2 + 4 + ... + 2^k or 1 + 3 + 9 + ... patterns.

✅ Solved examples

1. Find the sum of the first 6 terms of 2, 6, 18, ...
a = 2, r = 3. S_6 = 2(3^6 − 1)/(3 − 1) = 2(729 − 1)/2 = 728.
2. Sum the GP 1 + 2 + 4 + ... + 512.
a = 1, r = 2; 512 = 2^9 is the 10th term. S_10 = (2^10 − 1)/(2 − 1) = 1024 − 1 = 1023.
3. Find 3 + 1 + 1/3 + ... to 5 terms.
a = 3, r = 1/3. S_5 = 3(1 − (1/3)^5)/(1 − 1/3) = 3(1 − 1/243)/(2/3) = (9/2)(242/243) = 1089/243 = 121/27.
4. The sum of the first n terms of 5, 10, 20, ... is 1275. Find n.
S_n = 5(2^n − 1)/(2 − 1) = 5(2^n − 1) = 1275 ⇒ 2^n − 1 = 255 ⇒ 2^n = 256 = 2^8 ⇒ n = 8.

✏️ Practice — try these, take hints as needed

1. Sum the first 5 terms of 4, 12, 36, ...
a = 4, r = 3.
S_5 = 4(3^5 − 1)/2.
3^5 = 243.
484
2. Find 1 + 3 + 9 + ... + 729.
729 = 3^6, the 7th term.
S_7 = (3^7 − 1)/2.
3^7 = 2187.
1093
3. Sum of first n terms of 3, 6, 12, ... is 765. Find n.
S_n = 3(2^n − 1).
2^n − 1 = 255.
2^n = 256.
8
4. Sum the GP 2 − 4 + 8 − 16 to 6 terms.
r = −2.
S_6 = 2((−2)^6 − 1)/(−2 − 1).
(−2)^6 = 64.
−42
5. Find 6 + 2 + 2/3 + ... to 4 terms.
a = 6, r = 1/3.
S_4 = 6(1 − (1/3)^4)/(2/3).
(1/3)^4 = 1/81.
720/81 = 80/9

📝 Topic test — 8 questions

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