Question

Which term of the $G.P.2,8,32,......$ upto ${}^{\prime }{n}^{\prime }$ terms is $131072$ ? Find the sum of $6$ terms.

Answer 1

$\begin{array}{c}2,8,32,.........\\ a=2,r=\frac{8}{2}=4\\ a_n=a{r}^{n-1}\\ 131072=2{\left(4\right)}^{r-1}\\ 65536={\left(4\right)}^{n-1}\\ 4^8={\left(4\right)}^{n-1}\\ n-1=8\\ n=8+1\\ n=9\\ S_n=a\frac{\left(r^n-1\right)}{r-1}ifr>1\\ S_6=\frac{2\left(4^6-1\right)}{4-1}\\ S_6=\frac{2\left(4096-1\right)}{3}=\frac{2\times 4095}{3}=2730\end{array}$