Question

If in a G.P., 5th term and the 12th term are 9 and $\frac{1}{243}$ respectively, find the 9th term of G.P.

• A. $\frac{1}{7}$
• B. $\frac{1}{8}$
• C. $\frac{1}{9}$
• D. $\frac{1}{81}$

Answer 1

The correct option is C $\frac{1}{9}$

Let $a$ be the first term and $r$ be the common ratio of the GP.
For a GP, the $nth$ term is given by $T_n=a{r}^{n-1}$
Given, $T_5=9$
$=>a{r}^{5-1}=9$
$=>a{r}^4=9$ -- (1)
And,
$T_{12}=\frac{1}{243}$
$=>a{r}^{12-1}=\frac{1}{243}$
$=>a{r}^{11}=\frac{1}{243}$ -- (2)
Dividing eqn 2 by eqn 1, we get
$=>r^7=\frac{1}{243\times 9}=\frac{1}{3^5\times 3^2}=\frac{1}{3^7}$
$=>r=\frac{1}{3}$
Substituting in eqn $1$, we get
$a\times (\frac{1}{3}{)}^4=9$
$a=729$
And $T_9=a{r}^{9-1}=729\times (\frac{1}{3}{)}^8=\frac{1}{9}$