Question

In a GP the $3^{rd}$ term is $24$ and the $6^{th}$ term is $192$. Find the ${10}^{th}$ term.

Answer 1

Let a be the first term and r be the common ratio.
It is given that $a_3=24and{a}_6=192$
As we know that, $a_n=a{r}^{n-1}$

$\Rightarrow a_3=a{r}^{2}and{a}_6=a{r}^5$

$\Rightarrow a{r}^2=24anda{r}^5=192$

$\Rightarrow \frac{a{r}^5}{a{r}^2}=\frac{192}{24}$

$\Rightarrow r^3=8$

$\Rightarrow r^3=2^3$

$\Rightarrow r=2$

Putting $r=2ina{r}^2=24$ then we get,

$\Rightarrow a{\left(2\right)}^2=24$

$\Rightarrow 4a=24$

$\Rightarrow a=6$

Now, ${10}^{th}term=a_{10}=a{r}^9\left(i\right)$
Putting $a=6$ and $r=2$ in (i) then we get,

$\Rightarrow a_{10}=\left(6\right){\left(2\right)}^9=3072$