Sequences Series

For any three positive real numbers a, b and c, 9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c). Then

  1. b, c and a are in G.P.
  2. b, c and a are in A.P.
  3. a, b and c are in A.P.
  4. a, b and c are in G.P.

Solution

(15a – 3b)2 + (15a – 5c)2 + (3b – 5c)2 = 0

Let, 15a = 3b = 5c = 45λ

a = 3λ; b = 15λ; c = 9λ

2c = a + b

b, c, a are in A.P.

The correct option is B.