Find the third term from the end of the G.P.

$\dfrac{2}{27}, \dfrac{2}{9}, \dfrac{2}{3}, …….., 162.$

Common ratio,

$\dfrac{\dfrac{2}{9}}{\dfrac{2}{27}} = \dfrac{2 \times 27}{2 \times 9} = 3$.

So, the above sequence is a G.P. with r = 3.

Let there be n terms.

$\Rightarrow ar^{n - 1} = 162 \\[1em] \Rightarrow \dfrac{2}{27} \times (3)^{n - 1} = 162 \\[1em] \Rightarrow \dfrac{2}{3^3} \times 3^{n - 1} = 162 \\[1em] \Rightarrow 3^{n - 1 - 3} = \dfrac{162}{2} \\[1em] \Rightarrow 3^{n - 4} = 81 \\[1em] \Rightarrow 3^{n - 4} = 3^4 \\[1em] \Rightarrow n - 4 = 4 \\[1em] \Rightarrow n = 8.$

m^th term from end is (n - m + 1)^th term from starting.

So, 3^rd term from end is (8 - 3 + 1) = 6^th term from starting.

a₆ = ar⁵

= $\dfrac{2}{27} \times (3)^5$

= $\dfrac{2}{27} \times 243$

= 2 × 9

= 18.

Hence, 3^rd term from end is 18.