Question

Which term of the G.P. 3, 3√3, 9, ...... is 2187 ?

(a) 13 

(b) 14 

(c) 15 

(d) 16

Answer 1

(a) 13 

Here first term a = 3, common ratio r = \(\frac{3\sqrt3}{3}\) = \(\sqrt3\)

Let the nth term be 2187. Then, 

T_n = ar^{n – 1} = 2187

⇒ 3 x (\(\sqrt3\))^{n - 1} = 2187

⇒ (\(\sqrt3\))^{n - 1} = 729

⇒ (\(\sqrt3\))^{n - 1} = 3⁶ = (\(\sqrt3\))¹²

⇒ n – 1 = 12 ⇒ n = 13.