If a, b, c are in A.P., prove that:

a2 + c2 + 4ac = 2 (ab + bc + ca)

a2 + c2 + 4ac = 2 (ab + bc + ca)


a2 + c2 + 2ac - 2ab - 2bc = 0


(a + c - b)2 – b2 = 0


a + c - b = b


a + c - 2b = 0


Since a, b, c are in AP


b - a = c - b


a + c - 2b = 0


Hence proved


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