How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461 ?

Let n terms be added.

Common ratio = $\dfrac{4}{1}$ = 4.

$\Rightarrow S = \dfrac{a(r^n - 1)}{(r - 1)} ……….(As |r| \gt 1)\\[1em] \Rightarrow 5461 = \dfrac{1[(4)^n - 1]}{4 - 1} \\[1em] \Rightarrow \dfrac{4^n - 1}{3} = 5461 \\[1em] \Rightarrow 4^n - 1 = 16383 \\[1em] \Rightarrow 4^n = 16384 \\[1em] \Rightarrow 4^n = 4^7 \\[1em] \Rightarrow n = 7.$

Hence, 7 terms must be added to get a sum of 5461.