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

$\begin{array}{1 1}3 \\ 4\\ 5\\ 6 \end{array}$

## 1 Answer

Toolbox:
• Sum of $n$ terms of a G.P.=$S_n=a.\large\frac{r^n-1}{r-1}$
Given that sum of $n$ terms of the G.P.$3+3^2+3^3+.....=120$
$\Rightarrow\:$ First term$=a=3$, common ratio$=r=\large\frac{3^2}{3}$$=3 and S_n=120 We know that sum of n terms of a G.P.=S_n=a.\large\frac{r^n-1}{r-1} \Rightarrow\:120=3.\large\frac{3^n-1}{3-1} \Rightarrow\:\large\frac{120\times 2}{3}$$=3^n-1$
$\Rightarrow\:80=3^n-1$
$\Rightarrow\:3^n=81=3^4$
Since the base on either sides are same, power on either sides are equal
$\Rightarrow\:n=4$
$i.e.,$ $4$ terms are required to give the sum of the G.P.$=120$
answered Feb 26, 2014

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