Question

How many terms of G.P. $3,3^2,3^3$, … are needed to give the sum 120?

Answer 1

The given G.P. is $3,3^2,3^3,$ …

Let n terms of this G.P. be required to obtain the sum as 120.

Sn = $a\frac{(r^n-1)}{r-1}$

Here, a = 3 and r = 3

∴ Sn = 120 = $3\frac{(3^n-1)}{3-1}$

⇒ 120 = $3\frac{(3^n-1)}{2}$

⇒ $\frac{120\times2}{3}=3^n-1$

⇒ $3^n-1$ = 80

⇒ $3^n$ = 80

⇒ $3^n$ =$3^4$

∴ n = 4

Thus, four terms of the given G.P. are required to obtain the sum as 120.