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,
$3,1,\frac{1}{3},\dots$
which is a $G.P$. With
$a=3,r=\frac{1}{3}$
$\therefore a_6=a{r}^5\left[\because a_n=a{r}^{n-1}\right]$
$\Rightarrow a_6=3{\left(\frac{1}{3}\right)}^5$
$=3\times \frac{1}{3^5}$
$=\frac{1}{3^4}$
$=\frac{1}{81}$