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

The number of ways of selecting $3$ numbers that are in $G.P.$ from the set $\{1, 2, 3, \ldots, 100\}$ is

If $64, 27, 36$ are the $P^{\text{th}}$, $Q^{\text{th}}$, and $R^{\text{th}}$ terms of a $GP$, then $P+2Q$ is equal to

Insert two numbers between $3$ and $81$ so that the resulting sequence is a $G.P.$

If $S$ is the sum to infinity of a $G.P.$,whose first term is $a$,then the sum of the first $n$ terms is

$A$ manufacturer reckons that the value of a machine,which costs him $Rs. 15625$,will depreciate each year by $20\%$. Find the estimated value at the end of $5$ years.

Vedclass Test Series

Exam Paper Generator

Online Exam Module