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Given a G.P. with $a=729$ and $7^{th}$ term is $64$. Find $S_7.$

$\begin{array}{1 1}2059 \\ 665 \\ \large\frac{2059}{3} \\ 1995 \end{array} $

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  • $n^{th}$ term of a G.P.= $t_n=a.r^{n-1}$
Given that in a G.P.first term $a=729$,and $7^{th}$ term $=t_7=64$
We know that $t_n=a.r^{n-1}$
$\therefore t_7=64=a.r^{7-1}$
$\Rightarrow\:64=729\times r^6$
But $64=2^6$ and $729=3^6$
We know that Sum of $n$ terms of a G.P.=$S_n=a.\large\frac{1-r^n}{1-r}$
$\therefore S_7=729.\large\frac{1-(2/3)^7}{1-2/3}$
$=729\times 3\times \big(1-(\large\frac{2}{3})^7\big)$
$=3^7\times\large\frac{ (3^7-2^7)}{3^7}$
answered Feb 27, 2014 by rvidyagovindarajan_1

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