7 min readConcept sheetShort calculation; 10-mark growth or depreciation problem

Geometric progressions (GP)

In a GP each term is multiplied by the same common ratio r. Growth by a fixed percentage, such as sales rising 10% a year, is a GP.

Common ratior = T₂ ÷ T₁

nth termTₙ = arⁿ⁻¹

Sum of n termsSₙ = a(rⁿ − 1) ÷ (r − 1), r ≠ 1

Sum to infinityS∞ = a ÷ (1 − r), −1 < r < 1

GP formulas

Sequencea, ar, ar², ar³, …

nth termTₙ = arⁿ⁻¹

Sum, r > 1Sₙ = a(rⁿ − 1) ÷ (r − 1)

Sum, r < 1Sₙ = a(1 − rⁿ) ÷ (1 − r)

Sum to infinity (−1 < r < 1)S∞ = a ÷ (1 − r)

Growth at g% a yearr = 1 + g/100

Decline at g% a yearr = 1 − g/100

AP or GP?

AP: addEach step adds the same amount, e.g. a Rs. 2,500 increment. Growth is a straight line.

GP: multiplyEach step multiplies by the same ratio, e.g. 10% growth (r = 1.1). Growth curves upward.

A term and a sum

Problem

For the GP 3, 6, 12, …, find the 8th term and the sum of the first 8 terms.

Working

a = 3, r = 6 ÷ 3 = 2, n = 8.
T8 = 3 × 2⁷ = 3 × 128 = 384.
S8 = 3(2⁸ − 1) ÷ (2 − 1) = 3 × 255 = 765.

Answer

T8 = 384 and S8 = 765.

Sales growth

Problem

A Kandy bakery sold Rs. 400,000 in year 1. Sales grow 10% a year. Find (a) sales in year 5 and (b) total sales over the 5 years.

Working

a = 400,000, r = 1.1.
(a) T5 = 400,000 × 1.1⁴ = 400,000 × 1.4641 = 585,640.
(b) S5 = 400,000 × (1.1⁵ − 1) ÷ (1.1 − 1).
1.1⁵ = 1.61051, so S5 = 400,000 × 0.61051 ÷ 0.1 = 400,000 × 6.1051 = 2,442,040.

Answer

(a) Rs. 585,640 (b) Rs. 2,442,040

Reducing-balance depreciation and the multiplier

Problem

(a) A Rs. 1,000,000 machine loses 20% of its value each year. Find its value after 3 years. (b) Rs. 1,000 of new spending is followed by rounds in which 60% of the last round is spent again. Find the total spending.

Working

(a) r = 1 − 0.20 = 0.8. Value after 3 years = 1,000,000 × 0.8³ = 1,000,000 × 0.512 = 512,000.
(b) Rounds: 1,000 + 600 + 360 + …, a GP with a = 1,000 and r = 0.6.
S∞ = 1,000 ÷ (1 − 0.6) = 1,000 ÷ 0.4 = 2,500.

Answer

(a) Rs. 512,000 (b) Rs. 2,500 in total

In Sri Lankan business

Growth at a fixed percentage (sales, prices, population) is a GP. So is reducing-balance depreciation of a vehicle in a Sri Lankan company's books: each year's value is the last year's times a ratio below 1.

Exam angle

Short answer: “Find the 6th term of 2, 6, 18, …” 2 × 3⁵ = 486.

Structured question: “Find the sum to infinity of 12 + 6 + 3 + … and explain why it exists.” r = 1/2 lies between −1 and 1, so S∞ = 12 ÷ (1 − 1/2) = 24.

Essay: “Sales of Rs. 400,000 grow 10% a year. Find year-5 sales and 5-year total, and compare with growth of Rs. 40,000 a year.” Rs. 585,640 and Rs. 2,442,040; AP total Rs. 2,400,000, so GP is Rs. 42,040 higher.

Remember it

GP = Grows by multiplying. Percent growth g% means r = 1 + g/100; percent decline means r = 1 − g/100.