# 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.

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