Question

If 4a²+9b²+16c² = 2(3ab+6bc+4ca), where a, b, c are non–zero real numbers then a, b, c are in

(a)  A.P.

(b)  G.P.

(c)  H.P. 

(d)  None of these 

Answer 1

Correct option  (c) H.P

Explanation:

Multiply by 2 on both sides

4a²+4a²+9b²+9b²+16c²+16c²–12ab–24bc–16ca = 0

⇒ (2a–3b)²+(3b–4c)²+(4c–2a)²=0 

⇒ 2a = 3b = 4c = λ

⇒ a = λ/2, b = λ/3 , c = λ/4

2,3,4 are in AP

⇒ 1/2, 1/3, 1/4 are in H.P.

⇒ λ/2, λ/3,λ/4 are in HP gives

a, b, c are in HP