Question

Which term of the $GP\sqrt{3},3,3\sqrt{3},......$ is $729$?

Answer 1

Formula,

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

Given,

$a=\sqrt{3},r=\frac{3}{\sqrt{3}}=\sqrt{3}$

$729=\sqrt{3}(\sqrt{3}{)}^{n-1}$

$3^6=3^{\frac{n}{2}}$

$\therefore 6=\frac{n}{2}$

$\therefore n=12$

$12$th term is $729$.