The first term of a G.P. is 27 and 8th term is $\dfrac{1}{81}.$ Find the sum of its first 10 terms.

Given, a = 27 and $a_8 = \dfrac{1}{81}.$

By formula,

$a$n = ar^{n - 1} \\[1em] \therefore a8 = ar^7 \\[1em] \Rightarrow \dfrac{1}{81} = 27r^7 \\[1em] \Rightarrow \dfrac{1}{81 \times 27} = r^7 \\[1em] \Rightarrow \dfrac{1}{3^4 \times 3^3} = r^7 \\[1em] \Rightarrow r^7 = \dfrac{1}{3^7} \\[1em] \Rightarrow r = \dfrac{1}{3}.

By formula,

$S$n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore S{10} = \dfrac{27\Big[\Big(\dfrac{1}{3}\Big)^{10} - 1\Big]}{\dfrac{1}{3} - 1} \\[1em] = \dfrac{27(\dfrac{1}{3^{10}} - 1)}{\dfrac{1 - 3}{3}} \\[1em] = \dfrac{27 \dfrac{(1 - 3^{10})}{3^{10}}}{-\dfrac{2}{3}} \\[1em] = \dfrac{3^3 \dfrac{(1 - 3^{10})}{3^{10}}}{-2} \times 3 \\[1em] = \dfrac{3^4(1 - 3^{10})}{-2 \times 3^{10}} \\[1em] = \dfrac{3^4}{2} \times \dfrac{-(1 - 3^{10})}{3^{10}} \\[1em] = \dfrac{81}{2} \Big(\dfrac{3^{10} - 1}{3^{10}}\Big) \\[1em] = \dfrac{81}{2} \Big(1 - \dfrac{1}{3^{10}}\Big).

Hence, the sum of first 10 terms of G.P. is $\dfrac{81}{2} \Big(1 - \dfrac{1}{3^{10}}\Big).$