Question

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

• A. $12$
• B. $13$
• C. $14$
• D. $15$

Answer 1

Answer: (B)

$\left(13\right)$

Since, $3$, $\sqrt{3}$, $1$ are in G.P. and $a_n=\frac{1}{243}$
Here, first term $a=3$ and common ratio $r=\frac{1}{\sqrt{3}}$
Since, $a_n=a{r}^{n-1}$
$⟹a{r}^{n-1}=\frac{1}{243}$, As given $a_n=\frac{1}{243}$

$⟹3{\left(\frac{1}{\sqrt{3}}\right)}^{n-1}=\frac{1}{3^5}$
$⟹3^1\times 3^{-\frac{n-1}{2}}=3^{-5}$ [Since, $\frac{1}{a^m}=a^{-m}$]
$⟹3^{[1-\frac{(n-1)}{2}]}=3^{-5}$ [Since, $a^m\times a^n=a^{m+n}$]
When the bases are equal then their powers are equal.
$⟹\frac{2-n+1}{2}=-5$
$⟹3-n=-10$

$⟹-n=-10-3$
$⟹-n=-13\therefore n=13$