Question

For any three positive real numbers a, b and c,

9(25a² + b²) + 25(c² - 3ac) = 15b(3a + c) Then

(A)  b, c and a are in AP

(B)  a, b and c are in AP

(C)  a, b and c are in GP

(D)  b, c and a are in GP 

Answer 1

Correct option (A) b, c and a are in AP

We have

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

1/2 [(15a - 3b)² + (3b - 5c)² + (5c - 15a)²] = 0

It is possible when 15a = 3b = 5c.

Therefore, b = 5c/3 and a = c/3.

That is, a + b = 2c. 

Thus, b, c and a are in AP.