For any three positive real numbers a, b and c, 9(25a^{2} + b^{2}) + 25(c^{2} – 3ac) = 15b(3a + c). Then

- b, c and a are in G.P.
- b, c and a are in A.P.
- a, b and c are in A.P.
- 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.**