In a G.P, t2 + t5 = 216 and t4 : t6 = 1 : 4 and all the terms are inte

1.	In a G.P, t2 + t5 = 216 and t4 : t6 = 1 : 4 and all the terms are integers, then its first term is(a) 16 (b) 14 (c) 12 (d) None of these
Answer» (c) 12 Let a and r be the first term and common ratio respectively of the given G.P. Then, t₂ = ar, t₄ = ar³, t₅ = ar⁴, t₆ = ar⁵ Given, t₂ + t₅ = ar + ar⁴ = 216 ...(i) and \(\frac{t_4}{t_6}\) = \(\frac{ar^3}{ar^5}\) = \(\frac{1}{4}\) ⇒ \(\frac{1}{r^2}\) = \(\frac{1}{4}\) ⇒ \(\frac{1}{r}\) = \(\frac{1}{2}\) ...(ii) Now putting r = 2, in (i) we get 2a + 16a = 216 ⇒ 18a = 216 ⇒ a = 12.

In a G.P, t2 + t5 = 216 and t4 : t6 = 1 : 4 and all the terms are integers, then its first term is(a) 16 (b) 14 (c) 12 (d) None of these

Answer»

(c) 12

Let a and r be the first term and common ratio respectively of the given G.P. Then, t₂ = ar, t₄ = ar³, t₅ = ar⁴, t₆ = ar⁵

Given, t₂ + t₅ = ar + ar⁴ = 216 ...(i)

and \(\frac{t_4}{t_6}\) = \(\frac{ar^3}{ar^5}\) = \(\frac{1}{4}\) ⇒ \(\frac{1}{r^2}\) = \(\frac{1}{4}\) ⇒ \(\frac{1}{r}\) = \(\frac{1}{2}\) ...(ii)

Now putting r = 2, in (i) we get

2a + 16a = 216 ⇒ 18a = 216 ⇒ a = 12.