Given a G.P. with a=729 and 7^{text{th}} term | vedclass

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(A) Given $a=729$ and $a_{7}=64$.Let $r$ be the common ratio of the $G.P.$We know that $a_{n}=a r^{n-1}$.$a_{7}=a r^{6} = 729 r^{6} = 64$.$r^{6} = \fr

Vedclass · Accepted Answer

$2059$

Vedclass · Answer

(A) Given $a=729$ and $a_{7}=64$.Let $r$ be the common ratio of the $G.P.$We know that $a_{n}=a r^{n-1}$.$a_{7}=a r^{6} = 729 r^{6} = 64$.$r^{6} = \frac{64}{729} = \left(\frac{2}{3}ight)^{6}$.Thus,$r = \frac{2}{3}$.The sum of $n$ terms of a $G.P.$ is given by $S_{n}=\frac{a(1-r^{n})}{1-r}$.$S_{7}=\frac{729(1-(2/3)^{7})}{1-2/3} = \frac{729(1-(2/3)^{7})}{1/3}$.$S_{7} = 3 	imes 729 	imes \left(1 - \frac{2^{7}}{3^{7}}ight) = 3^{7} 	imes \left(\frac{3^{7}-2^{7}}{3^{7}}ight)$.$S_{7} = 3^{7} - 2^{7} = 2187 - 128 = 2059$.

Given a $G.P.$ with $a=729$ and $7^{\text{th}}$ term $64,$ determine $S_{7}$.

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