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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $S_{7}$.

Similar Questions

If $1 + \sin x + \sin^2 x + \dots \infty = 4 + 2\sqrt{3}$ for $0 < x < \pi$,then:

How many terms of the $G.P.$ $3, \frac{3}{2}, \frac{3}{4}, \ldots$ are needed to give the sum $\frac{3069}{512} ?$

If $1 + \cos \alpha + \cos^2 \alpha + \dots \infty = 2 - \sqrt{2}$,then $\alpha$ $(0 < \alpha < \pi)$ is

The sum of an infinite geometric progression is $\frac{4}{3}$ and the first term is $\frac{3}{4}$. The common ratio is

If $4^{\text{th}}$,$10^{\text{th}}$,and $16^{\text{th}}$ terms of a $G$.$P$. are $x, y$,and $z$ respectively,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module