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

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

 

 

 

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