Question

If $6th$ term of a G.P is $-\frac{1}{32}$, and $9th$ term is $\frac{1}{256}$, then the $11th$ term is

• A. $1024$
• B. $1/1024$
• C. $1/256$
• D. $1/512$

Answer 1

The correct option is B $1/1024$

Given, $T_6=a{r}^{6-1}=-\frac{1}{32}$

$\Rightarrow a{r}^5=-\frac{1}{32}$ ........$\left(1\right)$

$T_9=a{r}^{9-1}=\frac{1}{256}$

$\Rightarrow a{r}^8=\frac{1}{256}$ ........$\left(2\right)$

Eqn$\left(2\right)$ divided by Eqn$\left(1\right)$

$\Rightarrow \frac{a{r}^8}{a{r}^5}=\frac{1}{256}\times \frac{-32}{1}$

$\Rightarrow r^{8-5}=\frac{-32}{256}=-\frac{1}{8}$

$\Rightarrow r^3=-\frac{1}{8}$

$\therefore r=-\frac{1}{2}$

From $\left(1\right)$ we have $a{r}^5=-\frac{1}{32}$

$\Rightarrow a\times {(-\frac{1}{2})}^5=-\frac{1}{32}$

$\Rightarrow a\times \frac{1}{32}=\frac{1}{32}$

$\therefore a=1$

Now $T_{11}=a{r}^{11-1}=a{r}^{10}=1\times {(-\frac{1}{2})}^{10}=\frac{1}{1024}$

Hence the ${11}^{th}$ term is $\frac{1}{1024}$