Question

Find the sum of the following geometric series :
(i) 0.15 + 0.015 + 0.0015 + … to 8 terms;

(ii) √2 + 1/√2 + 1/2√2 + …. to 8 terms;

(iii) 2/9 – 1/3 + 1/2 – 3/4 + … to 5 terms;

Answer 1

(i) 0.15 + 0.015 + 0.0015 + … to 8 terms

Given:

a = 0.15

r = t₂/t₁ = 0.015/0.15 = 0.1 = 1/10

n = 8

By using the formula,

Sum of GP for n terms = a(1 – rⁿ )/(1 – r)

a(1 – rⁿ )/(1 – r) = 0.15 (1 – (1/10)⁸) / (1 – (1/10))

= 0.15 (1 – 1/10⁸) / (1/10)

= 1/6 (1 – 1/10⁸)

(ii) √2 + 1/√2 + 1/2√2 + …. to 8 terms;

Given:

a = √2

r = t₂/t₁ = (1/√2)/√2 = 1/2

n = 8

By using the formula,

Sum of GP for n terms = a(1 – rⁿ )/(1 – r)

a(1 – rⁿ )/(1 – r) = √2 (1 – (1/2)⁸) / (1 – (1/2))

= √2 (1 – 1/256) / (1/2)

= √2 ((256 – 1)/256) × 2

= √2 (255 × 2)/256

= (255√2)/128

(iii) 2/9 – 1/3 + 1/2 – 3/4 + … to 5 terms;

Given:

a = 2/9

r = t₂/t₁ = (-1/3) / (2/9) = -3/2

n = 5

By using the formula,

Sum of GP for n terms = a(1 – rⁿ )/(1 – r)

a(1 – rⁿ )/(1 – r) = (2/9) (1 – (-3/2)⁵) / (1 – (-3/2))

= (2/9) (1 + (3/2)⁵) / (1 + 3/2)

= (2/9) (1 + (3/2)⁵) / (5/2)

= (2/9) (1 + 243/32) / (5/2)

= (2/9) ((32 + 243)/32) / (5/2)

= (2/9) (275/32) × 2/5

= 55/72