9 min readConcept sheetPart II (2 marks); sums like 7 + 77 + 777 + ... and infinite GP word problems are Part III (5 marks)

Sum of a geometric series, finite and infinite

Finite GP: Sₙ = a(rⁿ − 1)/(r − 1), r ≠ 1. Infinite GP with −1 < r < 1: S∞ = a/(1 − r).

Finite sumSₙ = a(rⁿ − 1)/(r − 1) for r ≠ 1; Sₙ = na when r = 1

Same formula, handy when r < 1Sₙ = a(1 − rⁿ)/(1 − r)

Infinite sumS∞ = a/(1 − r), only when −1 < r < 1

Recurring decimals0.777... = 7/10 + 7/100 + ... = 7/9

Why the formula works

Write Sₙ = a + ar + ar² + ... + arⁿ⁻¹. Multiply by r: rSₙ = ar + ar² + ... + arⁿ. Subtract the first line from the second. All the middle terms cancel, leaving rSₙ − Sₙ = arⁿ − a, so Sₙ = a(rⁿ − 1)/(r − 1). When −1 < r < 1, rⁿ gets closer and closer to 0 as n grows, so the sum settles at a/(1 − r).

Example 1: sum of a finite GP

Problem

Find the sum of the first 8 terms of the GP 1, −3, 9, −27, ....

Working

a = 1, r = −3, n = 8.
S₈ = a(r⁸ − 1)/(r − 1) = [(−3)⁸ − 1]/(−3 − 1)
(−3)⁸ = 6561, so S₈ = (6561 − 1)/(−4) = 6560/(−4)
= −1640.

Answer

S₈ = −1640

Example 2: a word problem

Problem

On day 1, Arun sends a Deepavali greeting to 3 friends. Each person who receives it sends it to 3 new people the next day. How many people receive it in the first 7 days?

Working

Receivers each day: 3, 9, 27, ...: a GP with a = 3, r = 3, n = 7.
Formula: Sₙ = a(rⁿ − 1)/(r − 1).
S₇ = 3(3⁷ − 1)/(3 − 1)
3⁷ = 2187, so S₇ = 3 × 2186/2 = 3 × 1093
= 3279.
Check: 3 + 9 + 27 + 81 + 243 + 729 + 2187 = 3279.

Answer

3279 people receive the greeting.

Example 3: the sum 7 + 77 + 777 + ...

Problem

Find the sum to n terms of 7 + 77 + 777 + ....

Working

Take out 7: Sₙ = 7(1 + 11 + 111 + ... to n terms).
Multiply and divide by 9: Sₙ = (7/9)(9 + 99 + 999 + ...).
Write each as a power of 10 minus 1: Sₙ = (7/9)[(10 − 1) + (10² − 1) + (10³ − 1) + ...].
= (7/9)[(10 + 10² + ... + 10ⁿ) − n]
10 + 10² + ... + 10ⁿ is a GP with a = 10, r = 10, so its sum is 10(10ⁿ − 1)/9.
Sₙ = (7/9)[10(10ⁿ − 1)/9 − n] = 70(10ⁿ − 1)/81 − 7n/9.
Check n = 2: 70 × 99/81 − 14/9 = 84 = 7 + 77.

Answer

Sₙ = 70(10ⁿ − 1)/81 − 7n/9

Example 4: an infinite GP and a recurring decimal

Problem

(i) Find the sum of 24 + 12 + 6 + ... to infinity. (ii) Write 0.272727... as a fraction.

Working

(i) a = 24 and r = 1/2, which lies between −1 and 1.
S∞ = a/(1 − r) = 24/(1 − 1/2) = 24 ÷ 1/2 = 48.
(ii) 0.272727... = 27/100 + 27/10000 + ...: a GP with a = 27/100 and r = 1/100.
S∞ = (27/100)/(1 − 1/100) = (27/100)/(99/100) = 27/99 = 3/11.

Answer

(i) 48 (ii) 0.272727... = 3/11

In real life

A deposit earning compound interest grows as a GP, so the total of equal yearly deposits is a geometric sum. A festival forward sent on to 3 new people at each step reaches 3 + 9 + 27 + ... people very fast.

Exam angle

2 marks: “Find the sum of the infinite GP 16 + 4 + 1 + ....” a = 16, r = 1/4, and −1 < r < 1. S∞ = 16/(1 − 1/4) = 16 ÷ 3/4 = 64/3.

2 marks: “How many terms of 1 + 4 + 16 + ... make 341?” Sₙ = (4ⁿ − 1)/3 = 341, so 4ⁿ = 1024 = 4⁵ and n = 5.

5 marks: “Find the sum to n terms of 0.8 + 0.88 + 0.888 + ....” Sₙ = (8/9)(0.9 + 0.99 + 0.999 + ...) = (8/9)[n − (0.1 + 0.01 + ... to n terms)] = (8/9)[n − (1/9)(1 − (1/10)ⁿ)] = 8n/9 − (8/81)(1 − (1/10)ⁿ).

Remember it

Multiply by r, subtract, and the middle cancels. An infinite sum needs −1 < r < 1.