Question

prove that a minus b whole square a square + b square a + b whole square are consecutive terms of ap

Answer 1

Answers :-

if (a-b)^2 , (a^2+b^2) and (a + b)^2 are in A.p

then,

(a^2+ b^2) — (a–b)^2 = (a+b)^2 – (a^2 + b^2)

a^2 + b^2 – (a^2 + b^2 –2ab) = a^2 + b^2 + 2ab – (a^2 +b^2)

a^2 +b^2 – a^2 – b^2 +2ab = a^2 +b^2 +2ab –a^2 –b^2

after the calculation ,

2ab = 2ab

L.H.S = R.H.S

SO,

(a–b)^2 , (a^2+b^2) , (a+b)^2 are in AP with common difference 2ab.


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Answer 2

hope it helps you

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