Question

The $n^{\text{th}}$ term of the series $3,\sqrt{3},1,\dots$ is $\frac{1}{243},$ then $n=$

Answer 1

Clearly, the given series in G.P.
So, $3\cdot {\left(\frac{1}{\sqrt{3}}\right)}^{n-1}=\frac{1}{243}\Rightarrow n=13$