Question

Find the ${10}^{th}$ term of a $GP$ whose $8^{th}$ term is $192$ and the common ratio is $2$.

Answer 1

Using formula, for the nth term of a GP is

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

Given,

$r=2$

$a_8=a{r}^7\Rightarrow 192=a\times 2^7$

$\therefore a=\frac{3}{2}$

$a_{10}=a{r}^9$

$=\frac{3}{2}\times 2^9$

$=768$