Question

Which term of the following sequences

$\sqrt{3},3,3\sqrt{3},...$ is $729.$

Answer 1

The given sequence is $\sqrt{3},3,3\sqrt{3},...$

Clearly this sequence is in G.P

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

Let the $n^{th}$ term of the given sequence be $729$.

$a_n={ar}^{n-1}\Rightarrow {ar}^{n-1}=729$

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

$\Rightarrow {\left(3\right)}^{\frac{1}{2}}{\left(3\right)}^{\frac{n-1}{2}}={\left(3\right)}^6$

$\therefore \frac{1}{2}+\frac{n-1}{2}=6$

$\Rightarrow \frac{1+n-1}{2}=6$

$\Rightarrow n=12$

Thus, the ${12}^{th}$ term of the given sequence is $729$.