Question

If the $5^{th}$ term of a G.P. containing positive terms is $\frac{1}{3}$ and $9^{th}$ term is $\frac{16}{243}$, then the $4^{th}$ term will be

• A. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{2}{5}$

Answer 1

The correct option is B

$\frac{1}{2}$

$T_5$ = $a{r}^4$ = $\frac{1}{3}$ .......(i)

and $T_9$ = $a{r}^8$ = $\frac{16}{243}$ .......(ii)

Solving (i) and (ii), we get r = $\frac{2}{3}$ and a = $\frac{27}{16}$

Now $4^{th}$ term = $a{r}^3$ = $\frac{3^3}{2^4}.\frac{2^3}{3^3}$ = $\frac{1}{2}$