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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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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} $

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  • 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 by rvidyagovindarajan_1
 

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