The first term of a G.P. is 27. If the 8^th term be $\dfrac{1}{81}$, what will be the sum of 10 terms?

Given,

⇒ a = 27

⇒ a₈ = ar⁷ = $\dfrac{1}{81}$

⇒ 27r⁷ = $\dfrac{1}{81}$

⇒ r⁷ = $\dfrac{1}{81 \times 27}$

⇒ r⁷ = $\dfrac{1}{3^7}$

⇒ r⁷ = $\Big(\dfrac{1}{3}\Big)^7$

⇒ r = $\dfrac{1}{3}$.

Since, r < 1

$S = \dfrac{a(1 - r^n)}{(1 - r)} \\[1em] = \dfrac{27\Big[1 - \Big(\dfrac{1}{3}\Big)^{10}\Big]}{1 - \dfrac{1}{3}} \\[1em] = \dfrac{27\Big[1 - \Big(\dfrac{1}{3}\Big)^{10}\Big]}{ \dfrac{2}{3}} \\[1em] = \dfrac{81}{2}\Big[1 - \Big(\dfrac{1}{3}\Big)^{10}\Big] \\[1em] = \dfrac{81}{2}(1 - 3^{-10}).$

Hence, sum = $\dfrac{81}{2}(1 - 3^{-10}).$