Question

In a G.P. the $3^{rd}$ term is $24$ & $6^{th}$ term is $192$. Find the ${10}^{th}$ term.

Answer 1

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

$T_3=a{r}^2=24$...(1)

$T_6=a{r}^5=192$...(2)

From (1)

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

substitute value of a in (2)

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

$\Rightarrow r^3=8$

$\therefore r=2$

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

$T_{10}=a{r}^{10-1}$

$=\left(6\right)(2{)}^9$

$\therefore T_{10}=3072$