Question

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

Answer 1

Find the value of $n$:

Here, the GP is $\sqrt{3},3,3\sqrt{3},....$

First term $a=\sqrt{3}$,

Common ratio $r=\frac{3}{\sqrt{3}}=\sqrt{3}$ , $n=?$

Formula:

The general term of GP is $a_n=a{r}^{n-1}$.

Now, $a_n=\sqrt{3}\times {\left(\sqrt{3}\right)}^{n-1}=729$

$\begin{array}{cc}\Rightarrow & 3^{\frac{1}{2}+\frac{n}{2}-\frac{1}{2}}=3^6\\ \Rightarrow & \frac{n}{2}=6\\ \Rightarrow & n=12\end{array}$

Therefore, ${12}^{\text{th}}$ term of sequence is $729$.