Find the sum of first 10 terms of the G.P. 1, $\sqrt{3}$, 3, $3\sqrt{3}$, ………

Given,

a = 1

r = $\dfrac{\sqrt3}{1} = \sqrt3$

n = 10

We know that,

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

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

Substituting values we get :

$\Rightarrow S_{10} = \dfrac{1[(\sqrt3)^{10} - 1]}{\sqrt3 - 1} \\[1em] = \dfrac{(3)^{\dfrac{10}{2}} - 1}{\sqrt3 - 1} \\[1em] = \dfrac{(3)^{5} - 1}{\sqrt3 - 1} \\[1em] = \dfrac{243 - 1}{\sqrt3 - 1} \\[1em] = \dfrac{242}{\sqrt3 - 1}$

Rationalizing the Denominator :

$= \dfrac{242}{\sqrt3 - 1} \times \dfrac{\sqrt3 + 1}{\sqrt3 + 1}\\[1em] = \dfrac{242(\sqrt3 + 1)}{(\sqrt3)^2 - 1^2} \\[1em] = \dfrac{242(\sqrt3 + 1)}{3 - 1} \\[1em] = \dfrac{242(\sqrt3 + 1)}{2} \\[1em] = 121(\sqrt3 + 1).$

Hence, S₁₀ = $121(\sqrt3 + 1)$.