The 5th term from the end of the G.P. 2, 6, 18, …., 13122 is

1. 162

2. 486

3. 54

4. 1458

The above series is a G.P. with a = 2 and d = $\dfrac{6}{2} = 3.$

nth from end = $l\Big(\dfrac{1}{r}\Big)^{n - 1}$

∴ 5th term from end = $13122 \times \Big(\dfrac{1}{3}\Big)^{5 - 1}$

$= 13122 \times \dfrac{1}{3^4} \\[1em] = \dfrac{13122}{81} \\[1em] = 162.$

Hence, Option 1 is the correct option.