The 4^th and the 7^th terms of a G.P. are $\dfrac{1}{27} \text{ and } \dfrac{1}{729}$ respectively. Find the sum of n terms of this G.P.

Given,

$a_4 = \dfrac{1}{27}$

$ar^3 = \dfrac{1}{27}$ ………(i)

$a_7 = \dfrac{1}{729}$

$ar^6 = \dfrac{1}{729}$ ……..(ii)

Dividing (ii) by (i) we get,

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{\dfrac{1}{729}}{\dfrac{1}{27}} \\[1em] \Rightarrow r^3 = \dfrac{27}{729} \\[1em] \Rightarrow r^3 = \dfrac{1}{27} \\[1em] \Rightarrow r = \dfrac{1}{3}.$

Substituting value of r in (i) we get,

$\Rightarrow a\Big(\dfrac{1}{3}\Big)^3 = \dfrac{1}{27} \\[1em] \Rightarrow a \times \dfrac{1}{27} = \dfrac{1}{27} \\[1em] \Rightarrow a = 1.$

Since, r < 1

$S = \dfrac{a(1 - r^n)}{(1 - r)} \\[1em] = \dfrac{1\Big[1 - \Big(\dfrac{1}{3}\Big)^n \Big]}{1 - \dfrac{1}{3}} \\[1em] = \dfrac{\Big[1 - \Big(\dfrac{1}{3}\Big)^n \Big]}{\dfrac{2}{3}} \\[1em] = \dfrac{3}{2}\Big[1 - \Big(\dfrac{1}{3}\Big)^n \Big] \\[1em] = \dfrac{3}{2}\Big(1 - \dfrac{1}{3^n} \Big)$

Hence, sum = $\dfrac{3}{2}\Big(1 - \dfrac{1}{3^n} \Big).$