Find the 4th term from the end of the GP

Solution:

The given GP is $\frac{2}{27} \cdot \frac{2}{9}, \frac{2}{3}, \ldots, 162 . \rightarrow(1)$

The first term in the GP, $\mathrm{a}_{1}=\mathrm{a}=\frac{2}{27}$

The second term in the GP, $\mathrm{a}_{2}=\frac{2}{9}$

The common ratio, $r=3$

The last term in the given GP is $a_{n}=162$.

Second last term in the GP $=a_{n-1}=a r^{n-2}$

Starting from the end, the series forms another GP in the form,

$a r^{n-1}, a r^{n-2}, a r^{n-3} \ldots a r^{3}, a r^{2}, a r, a \rightarrow(2)$

Common ratio of this GP is $r^{\prime}=\frac{1}{r}$.

So, $r^{\prime}=\frac{1}{3}$

So, $4^{\text {th }}$ term of the GP $(2)$,

$a_{4}=a r^{3}$

$=162 \times \frac{1}{3^{3}}=6$

Hence, the $4^{\text {th }}$ term from the end of the given GP is 6 .