The first term of a G.P. is 27 and its 8th term is $\dfrac{1}{81}$. Find the sum of first seven terms of the G.P.

Given,

a = 27

$\Rightarrow T_8 = \dfrac{1}{81} \\[1em] \Rightarrow ar^{(8-1)} = \dfrac{1}{81} \\[1em] \Rightarrow ar^{7} = \dfrac{1}{81} \\[1em] \Rightarrow (27)r^{7} = \dfrac{1}{81} \\[1em] \Rightarrow r^{7} = \dfrac{1}{81 \times 27} \\[1em] \Rightarrow r^{7} = \dfrac{1}{3^4 \times 3^3} \\[1em] \Rightarrow r^{7} = \dfrac{1}{3^{4 + 3}} \\[1em] \Rightarrow r^{7} = \dfrac{1}{3^{7}} \\[1em] \Rightarrow r^{7} = \Big(\dfrac{1}{3}\Big)^7 \\[1em] \Rightarrow r = \dfrac{1}{3}.$

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 :

$\Rightarrow S_7 = \dfrac{27\Big[1 - \Big(\dfrac{1}{3}\Big)^7\Big]}{1 - \Big(\dfrac{1}{3}\Big)} \\[1em] = \dfrac{27\Big[1 - \Big(\dfrac{1}{2187}\Big)\Big]}{\Big(\dfrac{3 - 1}{3}\Big)} \\[1em] = \dfrac{27\Big(\dfrac{2187 - 1}{2187}\Big)}{\Big(\dfrac{3 - 1}{3}\Big)} \\[1em] = \dfrac{27\Big(\dfrac{2186}{2187}\Big)}{\Big(\dfrac{2}{3}\Big)} \\[1em] = \dfrac{27 \times 2186 \times 3}{2 \times 2187} \\[1em] = \dfrac{1093}{27}.$

Hence, S₇ = $\dfrac{1093}{27}$.