How many terms of the G.P. 1, 4, 16, … must be taken to have their sum equal to 341?

The above series is a G.P. with a = 1 and r = 4.

Let n terms be required to give the sum of 341.

By formula,

$S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore 341 = \dfrac{1(4^n - 1)}{4 - 1} \\[1em] \Rightarrow 341 = \dfrac{(4^n - 1)}{3} \\[1em] \Rightarrow 341 \times 3 = (4^n - 1) \\[1em] \Rightarrow 1023 = 4^n - 1 \\[1em] \Rightarrow 4^n = 1024 \\[1em] \Rightarrow 4^n = 4^5 \\[1em] \therefore n = 5.$

Hence, the required number of terms of the G.P. are 5.