find the least value of n for which the sum to n terms of the geometric series 1+0.98+(0.98)^2+(0.98)^3+...is greater than half the sum to infinity.

Question

6 marks question

Answer 1

a₁ = 1

r = 0.98

S = 1 / (1 - 0.98)

S = 50

S_n > 25

1(1 - 0.98ⁿ) / (1 - 0.98) > 25

(1 - 0.98ⁿ) / 0.02 > 25

1 - 0.98ⁿ > 0.5

-0.98ⁿ > -0.5

0.98ⁿ < 0.5

n ln 0.98 < ln 0.5

-0.02020 n < -0.6931

n > 43.314