Show That (a-b)2, (a2 +b 2) and (a+b)2 Are In A.P.


Given: (a-b)2, (a2+b2) and (a+b)2

The common differences, d1

⇒ (a+ b2) – (a – b)2

⇒(a2 + b2) – (a2 + b2 – 2ab)

⇒ a2 + b2 – a2 – b2 + 2ab


The common differences, d2

⇒ (a+b)2 – (a2 + b2)

⇒a2 + b2 + 2ab – a2} – b2


Therefore, d1 = d2

Hence it is proved that (a-b)2, (a2 +b2) and (a+b)2 are in A.P.

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