Question

Find the general term of $\sqrt{3},\frac{1}{\sqrt{3}},\frac{1}{3}\sqrt{3},\frac{1}{9}\sqrt{3},.......$

Answer 1

here first term $=\sqrt{3}$
and common ratio is $1/3$
so, $a_1\sqrt{3},r=\frac{1}{/}3$
so, nth term $a{r}^{n-1}$
so, an= $\left(\sqrt{3}\right){\left(1/3\right)}^{n-1}$