How many terms of the G.P. 3, 3², 3³, …. are needed to give the sum 120?

The given series is a G.P. with a = 3 and r = 3.

Let n terms be required to give the sum of 120.

By formula,

$S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore 120 = \dfrac{3(3^n - 1)}{3 - 1} \\[1em] \Rightarrow 120 = \dfrac{3(3^n - 1)}{2} \\[1em] \Rightarrow 240 = 3(3^n - 1) \\[1em] \Rightarrow 80 = 3^n - 1 \\[1em] \Rightarrow 81 = 3^n \\[1em] \Rightarrow 3^4 = 3^n \\[1em] \Rightarrow n = 4.$

Hence, the required number of terms of the G.P. are 4.