Solution:

Since a, b, c are in A.P so

b – a = c – b

b + b = c + a

2 b = c + a

Taking squares on both sides

(2 b) ² = (c + a)²

4b² = (c + a)²

Subtracting by 4ac on both sides, we get

4b² – 4ac = (c + a)² – 4 ac

4(b² – ac) = c² + a² + 2 ac – 4 ac

4(b² – ac) = c² + a² - 2 ac

4(b² – ac) = ( c – a )²