Question

Find the $4^{th}$ term from the end of the G.P. $\frac{2}{27},\frac{2}{9},\frac{2}{3},............,162.$

Answer 1

$G.P=\frac{2}{27},\frac{2}{9},\frac{2}{3},........,162$

Common ratio $=r=\frac{2}{27}\times \frac{9}{2}=3$

To find the $4^{th}$ term from the end, we consider the reversed $GP$ such that $162,\mathrm{54........}\frac{2}{3},\frac{2}{9},\frac{2}{27}$

Also $r=\frac{1}{3}$.

Now, for the $reversedGP$

$a=162,r=\frac{1}{3}$, $T_4=?$

As we know,

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

$T_4=162\times {\left(\frac{1}{3}\right)}^{3-1}$

$T_4=162\times \left(\frac{1}{9}\right)$

$T_4=18.$

Hence, $4^{th}$ term from the end of the GP will be $18$