The 4th and the 7th terms of a G.P. are $\dfrac{1}{27}$ and $\dfrac{1}{729}$ respectively. Find the sum of first 6 terms of the G.P.

Given,

⇒ T₄ = $\dfrac{1}{27}$

⇒ ar³ = $\dfrac{1}{27}$ …..(1)

Given,

⇒ T₇ = $\dfrac{1}{729}$

⇒ ar⁶ = $\dfrac{1}{729}$ ……….(2)

Dividing Equation (2) by Equation (1) :

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

Substituting, $r = \dfrac{1}{3}$ in equation 1 :

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

We know that,

The sum of the first n terms of a G.P. is given by:

$S_n = \dfrac{a(1 - r^n)}{1 - r}$ [For r < 1]

Substituting values we get :

$\Rightarrow S_6 = \dfrac{1\Big[1 - \Big(\dfrac{1}{3}\Big)^6\Big]}{1 - \Big(\dfrac{1}{3}\Big)} \\[1em] = \dfrac{\Big(1 - \dfrac{1}{729}\Big)}{\dfrac{3 - 1}{3}} \\[1em] = \dfrac{\Big(\dfrac{729 - 1}{729}\Big)}{\Big(\dfrac{3 - 1}{3}\Big)} \\[1em] = \dfrac{\Big(\dfrac{728}{729}\Big)}{\Big(\dfrac{2}{3}\Big)} \\[1em] = \dfrac{728}{729} \times \dfrac{3}{2} \\[1em] = \dfrac{364}{243}.$

Hence, S₆ = $\dfrac{364}{243}$.