Question

Which term of the G.P. $2,2\sqrt{2},4,...........$ is 32?

Answer 1

We can see that: $a=2,r=\frac{T_2}{T_1}=\frac{2\sqrt{2}}{2}=\sqrt{2},T_n=32,n=?$
$T_n=a{r}^{n-1}$
$32=2(\sqrt{2}{)}^{n-1}$
$\frac{32}{2}=(\sqrt{2}{)}^{n-1},16=(\sqrt{2}{)}^{n-1}$
$2^4=(\sqrt{2}{)}^{n-1}=(2^{1/2}{)}^{n-1}$ $[\because \sqrt{2}=2^{1/2}]$
$2^4=2^{\frac{n-1}{2}}$

$\therefore 4=\frac{n-1}{2}$
$8=n-1\Rightarrow n=8+1=9$
$\therefore$ 32 is the 9th term of the given G.P.