Question

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 1

(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.