Which term of the $GP.,$ $2,8,32, \ldots$ up to $n$ terms is $131072 ?$

Vedclass pdf generator app on play store
Vedclass iOS app on app store

Let $131072$ be the $n^{\text {th }}$ term of the given $G.P.$ Here $a=2$ and $r=4$

Therefore $\quad 131072=a_{n}=2(4)^{n-1} \quad$ or $\quad 65536=4^{n-1}$

This gives $4^{8}=4^{n-1}$

So that $n-1=8,$ i.e., $n=9 .$ Hence, $131072$ is the $9^{\text {th }}$ term of the $G.P.$

Similar Questions

Let for $n =1,2, \ldots \ldots, 50, S _{ a }$ be the sum of the infinite geometric progression whose first term is $n ^{2}$ and whose common ratio is $\frac{1}{(n+1)^{2}}$. Then the value of $\frac{1}{26}+\sum\limits_{n=1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)$ is equal to

  • [JEE MAIN 2022]

The first two terms of a geometric progression add up to $12.$ the sum of the third and the fourth terms is $48.$ If the terms of the geometric progression are alternately positive and negative, then the first term is

  • [AIEEE 2008]

If the sum of the series $1 + \frac{2}{x} + \frac{4}{{{x^2}}} + \frac{8}{{{x^3}}} + ....\infty $ is a finite number, then

The G.M. of the numbers $3,\,{3^2},\,{3^3},\,......,\,{3^n}$ is

If the product of three consecutive terms of $G.P.$ is $216$  and the sum of product of pair-wise is $156$, then the numbers will be