Which term of the G.P. 3, 3√3, 9, ...... is 2187 ?(a) 13 (b) 14 (c

1.	Which term of the G.P. 3, 3√3, 9, ...... is 2187 ?(a) 13 (b) 14 (c) 15 (d) 16
Answer» (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.

Which term of the G.P. 3, 3√3, 9, ...... is 2187 ?(a) 13 (b) 14 (c) 15 (d) 16

Answer»

(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.