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If 4a^2 + 9b^2 + 16c^2 = 2(3ab + 6bc + 4ca), where a, b, c are non–zero real numbers then a, b, c are in

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If 4a2 + 9b2 + 16c2 = 2(3ab + 6bc + 4ca), where a, b, c are non–zero real numbers then a, b, c are in

(a) A.P.

(b) G.P.

(c) H.P.

(d) None of these

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Correct option is (c) H.P. 

Multiply by 2 on both sides

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