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

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The given $G.P.$ is $3,3^{2}, 3^{3} \ldots$

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

$S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$

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 $120 .$

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