For the G.P. $\dfrac{1}{27}, \dfrac{1}{9}, \dfrac{1}{3}, ………, 81$;

find the product of fourth term from the beginning and the fourth term from the end.

Common ratio,

$\dfrac{\dfrac{1}{9}}{\dfrac{1}{27}} = \dfrac{27}{9} = 3$.

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

Let there be n terms.

$\Rightarrow ar^{n - 1} = 81 \\[1em] \Rightarrow \dfrac{1}{27} \times (3)^{n - 1} = 81 \\[1em] \Rightarrow 3^{n - 1} = 81 \times 27 \\[1em] \Rightarrow 3^{n - 1} = 3^4 \times 3^3 \\[1em] \Rightarrow 3^{n - 1} = 3^7 \\[1em] \Rightarrow n - 1 = 7 \\[1em] \Rightarrow n = 8.$

a₄ = ar³

= $\dfrac{1}{27} \times (3)^3$

= $\dfrac{1}{27} \times 27$

= 1.

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

So, 4^th term from end is (8 - 4 + 1) = 5^th term from starting.

a₅ = ar⁴

= $\dfrac{1}{27} \times (3)^4$

= $\dfrac{1}{27} \times 81$

= 3.

a₄.a₅ = 1 × 3 = 3.

Hence, product of fourth term from the beginning and the fourth term from the end = 3.