Which term of the $GP.,$ $2,8,32, \ldots$ up to $n$ terms is $131072 ?$
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 $G.P.$
if $x = \,\frac{4}{3}\, - \,\frac{{4x}}{9}\, + \,\,\frac{{4{x^2}}}{{27}}\, - \,\,.....\,\infty $ , then $x$ is equal to
The sum of first $20$ terms of the sequence $0.7,0.77,0.777, . . . $ is
Insert two numbers between $3$ and $81$ so that the resulting sequence is $G.P.$
Which term of the following sequences:
$\sqrt{3}, 3,3 \sqrt{3}, \ldots$ is $729 ?$
If $\frac{{x + y}}{2},\;y,\;\frac{{y + z}}{2}$ are in $H.P.$, then $x,\;y,\;z$ are in