Question

Which term of $GP$ $2,2\surd 2,\mathrm{4..}$ is $128$?

Answer 1

Which term of G.P $2,2\sqrt{2},\mathrm{4....}$ is 1289

$2,2\sqrt{2},4,.....,$

$a=2$

$r=\frac{2\sqrt{2}}{2}=\sqrt{2}$

$\to t_n=a{r}^{n-1}$

$\therefore t_n=128$

$\therefore a=2$

$r=\sqrt{2}$

$\to 128=2\times (\sqrt{2}{)}^{n-1}$

$\therefore \frac{128}{2}=(\sqrt{2}{)}^{n-1}$

$\therefore 64=(\sqrt{2}{)}^{n-1}$

$\therefore 2^6=(\sqrt{2}{)}^{n-1}$

$\therefore (\sqrt{2}.\sqrt{2}{)}^6=(\sqrt{2}{)}^{n-1}$

$\therefore {\sqrt{2}}^6=(\sqrt{2}{)}^{n-1}$

$\therefore (\sqrt{2}{)}^{12}=(\sqrt{2}{)}^{n-1}$

$\therefore n-1=12$

$\therefore n=13$