8 min readConcept sheetPart I (1 mark) and Part II (2 marks); three terms from their sum and product, or a growth or depreciation problem, is Part III (5 marks)

Geometric progression: nth term and three terms in GP

In a GP each term is the previous term times a fixed common ratio r (r ≠ 0). The nth term is tₙ = arⁿ⁻¹.

nth termtₙ = arⁿ⁻¹, with a ≠ 0 and r ≠ 0

Common ratior = t₂/t₁ = tₙ/tₙ₋₁

Three terms in GPa, b, c are in GP if b² = ac; take them as a/r, a, ar

Growth and decayGrowing by p% a year multiplies by (1 + p/100) each year; losing p% multiplies by (1 − p/100)

Key ideas

GP (geometric progression): a, ar, ar², ar³, ....
With a > 0: if r > 1 the terms grow fast; if 0 < r < 1 they shrink towards 0; if r < 0 the signs alternate.
Multiplying or dividing every term by the same non-zero number keeps a GP with the same r.
The reciprocals of the terms of a GP also form a GP, with common ratio 1/r.
For product problems, take three terms as a/r, a, ar, so the product is a³.

Example 1: a term and the position of a term

Problem

In the GP 3, 6, 12, 24, ..., find (i) the 9th term (ii) which term is 1536.

Working

a = 3 and r = 6/3 = 2.
Formula: tₙ = arⁿ⁻¹.
(i) t₉ = ar⁸ = 3 × 2⁸ = 3 × 256 = 768.
(ii) Put 3 × 2ⁿ⁻¹ = 1536, so 2ⁿ⁻¹ = 512 = 2⁹.
n − 1 = 9, so n = 10.

Answer

t₉ = 768, and 1536 is the 10th term.

Example 2: the condition for three terms

Problem

Find x so that x, x + 3 and x + 9 are three consecutive terms of a GP.

Working

For a GP, (middle term)² = product of the other two.
(x + 3)² = x(x + 9)
x² + 6x + 9 = x² + 9x
9 = 3x, so x = 3.
The terms are 3, 6, 12, with r = 2.

Answer

x = 3

Example 3: three terms from their product and sum

Problem

The product of three consecutive terms of a GP is 216 and their sum is 19. Find the terms.

Working

Let the terms be a/r, a, ar. Product: a³ = 216, so a = 6.
Sum: 6/r + 6 + 6r = 19, so 6/r + 6r = 13.
Multiply by r: 6 + 6r² = 13r, so 6r² − 13r + 6 = 0.
(2r − 3)(3r − 2) = 0, so r = 3/2 or r = 2/3.
r = 3/2 gives 4, 6, 9; r = 2/3 gives 9, 6, 4.

Answer

The terms are 4, 6 and 9.

Example 4: a depreciation word problem

Problem

A scooter bought for ₹ 80,000 loses 10% of its value every year. What is its value after 3 years?

Working

Each year the value is multiplied by 1 − 10/100 = 0.9.
Formula: tₙ = arⁿ⁻¹.
Values: ₹ 80,000 at the start, then 80,000 × 0.9, 80,000 × 0.9², ...: a GP with a = 80,000 and r = 0.9.
After 3 years the value is the 4th term: t₄ = 80,000 × (0.9)³.
(0.9)³ = 0.729, so t₄ = 80,000 × 0.729 = 58,320.

Answer

After 3 years the scooter is worth ₹ 58,320.

In real life

Lactic acid bacteria in warm milk keep doubling at roughly regular intervals, close to a GP, so a spoon of starter sets a bowl of curd overnight. A deposit earning compound interest also grows as a GP.

Exam angle

2 marks: “Find the 8th term of the GP 9, 3, 1, ....” a = 9, r = 1/3. t₈ = 9 × (1/3)⁷ = 9/2187 = 1/243.

2 marks: “Is 2, 6, 18, 54 a GP? If so, write its general term.” 6/2 = 18/6 = 54/18 = 3, so it is a GP with r = 3. tₙ = 2 × 3ⁿ⁻¹.

5 marks: “The 4th term of a GP is 54 and the 7th term is 1458. Find the GP.” ar³ = 54 and ar⁶ = 1458. Divide: r³ = 27, so r = 3. Then a = 54/27 = 2. The GP is 2, 6, 18, 54, .... Check: 2 × 3⁶ = 1458.

Remember it

AP adds, GP multiplies. In tₙ = arⁿ⁻¹ you multiply by r exactly n − 1 times.