Question

In a G.P., 3rd term is 16 and the 6th term is 128. Find the 10th term.

• A. $a_{10}=1048$
• B. $a_{10}=2048$
• C. $a_{10}=256$
• D. $a_{10}=512$

Answer 1

The correct option is B

$a_{10}=2048$

Given $a_3=16$ and $a_6=128$
$[\because a_n=a{r}^{n-1}]$
$\Rightarrow a{r}^2=16\dots \dots \left(i\right)$ and $a{r}^5=128\dots \dots \left(ii\right)$
Dividing $\left(ii\right)$ by $\left(i\right)$, we get $\frac{a{r}^5}{a{r}^2}=\frac{128}{16}$
$\Rightarrow r^3=8$
$r^3=2^3\Rightarrow r=2$
From $\left(i\right)$, we get $a(2{)}^2=16$
$\Rightarrow 4a=16$
$\Rightarrow a=4$
$\therefore a_{10}=a{r}^9=4(2{)}^9=4\times 512=2048$