Question

The sixth term in the sequence is $3,1, \frac{1}{3}$ is

• A. $\frac{1}{27}$
• B. $\frac{1}{9}$
• C. $\frac{1}{81}$
• D. $\frac{1}{17}$
• E. $\frac{1}{7}$

Answer 1

Answer: (C)

We have,
$3,1, \frac{1}{3}, \ldots$
which is a $G.P$. With
$ a=3, r=\frac{1}{3} $
$\therefore a_{6}=a r^{5} \,\,\,\left[\because a_{n}=a r^{n-1}\right]$
$\Rightarrow a_{6}=3\left(\frac{1}{3}\right)^{5}$
$=3 \times \frac{1}{3^{5}}$
$=\frac{1}{3^{4}}$
$=\frac{1}{81}$