How many terms of the G.P. 1, 4, 16, 64, …….. will make the sum 5461?

1. 6

2. 9

3. 7

4. 8

In the given G.P.,

a = 1

r = $\dfrac{4}{1}$ = 4

Formula for sum of n terms of a G.P.

$S_n = \dfrac{a(r^n-1)}{r-1}$ [r > 1]

Let the sum of n terms of the G.P. = 5461.

$\Rightarrow S_n = 5461 \\[1em] \Rightarrow 5461 = \dfrac{1(4^n - 1)}{(4 - 1)} \\[1em] \Rightarrow 5461 = \dfrac{(4^n - 1)}{3} \\[1em] \Rightarrow 5461 \times 3 = 4^n - 1 \\[1em] \Rightarrow 16383 = 4^n - 1 \\[1em] \Rightarrow 16383 + 1 = (2^2)^n \\[1em] \Rightarrow 16384 = 2^{2n} \\[1em] \Rightarrow 2^{14} = 2^{2n} \\[1em] \Rightarrow 2n = 14 \\[1em] \Rightarrow n = \dfrac{14}{2} \\[1em] \Rightarrow n = 7.$

Hence, option 3 is the correct option.