Question:
How many terms of G.P. $3,3^{2}, 3^{3}, \ldots$ are needed to give the sum $120 ?$
Solution:
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 .
$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$
Here, $a=3$ and $r=3$
$\therefore S_{n}=120=\frac{3\left(3^{n}-1\right)}{3-1}$
$\Rightarrow 120=\frac{3\left(3^{n}-1\right)}{2}$
$\Rightarrow \frac{120 \times 2}{3}=3^{n}-1$
$\Rightarrow 3^{n}-1=80$
$\Rightarrow 3^{n}=81$
$\Rightarrow 3^{n}=3^{4}$
$\therefore n=4$
Thus, four terms of the given G.P. are required to obtain the sum as 120