Question

In a $GP$ the $3^{rd}$ term is $24$ and $6^{th}$ term is $192$. Find the ${11}^{th}$ terms.

Answer 1

Formula,

$a_n=a{r}^{n-1}$

Given,

$a_3=a{r}^2=24$........(1)

$a_6=a{r}^5=192$.......(2)

$\left(2\right)÷\left(1\right)$ gives

$r^3=8\Rightarrow r=2$

$a{r}^2=24\Rightarrow a=\frac{24}{2^2}\Rightarrow a=6$

$a_{11}=a{r}^{10}=6\times 2^{10}=6144$