Find the seventh term from the end of the series :

$\sqrt{2}, 2, 2\sqrt{2}, ………., 32$

Since,

$\dfrac{2}{\sqrt{2}} = \dfrac{2\sqrt{2}}{2} = \sqrt{2}$.

So, the above sequence is a G.P. with r = $\sqrt{2}$.

Let there be n terms.

$\Rightarrow ar^{n - 1} = 32 \\[1em] \Rightarrow \sqrt{2} \times (\sqrt{2})^{n - 1} = 32 \\[1em] \Rightarrow \sqrt{2}^{n - 1 + 1} = 32 \\[1em] \Rightarrow (\sqrt{2})^n = (\sqrt{2})^{10} \\[1em] \Rightarrow n = 10.$

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

So, 7^th term from end is (10 - 7 + 1) = 4^th term from starting.

a₄ = ar³

= $\sqrt{2}(\sqrt{2})^3$

= $\sqrt{2} \times 2\sqrt{2}$

= 4.

Hence, 7^th term from end is 4.