Show that (a - b)2, (a2 + b2) and (a + b)2 are in AP.
Consider (a2 + b2) - (a - b)2
= (a2 + b2) - (a2 + b2 - 2ab)
= 2ab
Consider (a + b)2 - (a2 + b2)
= (a2 + b2 + 2ab) - (a2 + b2)
= 2ab
Since, the difference between consecutive terms is constant, therefore the terms are in AP.