Show that (a-b)², (a²+b²) , (a+b)² are in AP

1.	Show that (a-b)², (a²+b²) , (a+b)² are in AP
Answer» For being in AP common difference of successive terms of an AP be equal(a^2+b^2)-(a-b)^2=(a+b)^2-(a^2+b^2)LHS = (a^2+b^2)-(a-b)^2= a^2+b^2-a^2-b^2+2ab=2abRHS = (a+b)^2-(a^2+b^2)=a^2+b^2+2ab-a^2-b^2=2abTherefore, LHS=RHSHence, the common difference among successive terms is equal .So, the given terms are in AP.

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