Which term of the G.P. $\sqrt{3}, 3, 3\sqrt{3}$, 9, …… is 729 ?

Given,

The G.P. : $\sqrt{3}, 3, 3\sqrt{3}$, 9,……

a = $\sqrt{3}$

r = $\dfrac{3}{\sqrt3} = \sqrt3$

T_n = 729

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

$\Rightarrow 729 = \sqrt3 \times (\sqrt3)^{n - 1} \\[1em] \Rightarrow 729 = (\sqrt3)^{1 + n - 1} \\[1em] \Rightarrow 729 = (\sqrt3)^{n} \\[1em] \Rightarrow 3^6 = (3)^{\dfrac{n}{2}} \\[1em] \Rightarrow 6 = \dfrac{n}{2} \\[1em] \Rightarrow n = 6 \times 2 \\[1em] \Rightarrow n = 12.$

Hence, 12th term of G.P. is 729.