Solution:
We are given in the question that numbers a, b, c are in AP. So we haveb – a = c − b... (1)
⇒ 2b = a + c... (2)
⇒ b = … (3)
As given, a2, b2, c2 are in AP.
b2 − a2 = c2 − b2
⇒ (b − a) (b + a) = (c − b) (c + b)
We have b – a = c − b as a, b, c are assumed to be distinct numbers in AP. We use it and get,
b + a = c + b
⇒ a = c
So we have if a = 0 we have b = = a and b = = c.
So we have a = b = c.
So, the given statement is true.
