Question

Which term of the $G.P;2,2\sqrt{2},4,........$ is $128$?

Answer 1

We have,

G.P.$=2,2\sqrt{2},\mathrm{4.........}is128$

First term $a=2$

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

Number of terms $n=?$

Last term $l=128$

Then ,

We know that,

$l=a{r}^{n-1}$

$128=2{\left(\sqrt{2}\right)}^{n-1}$

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

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

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

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

$n-1=12$

$n=12+1$

$n=13$

Hence, this is the answer.