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# If A, B, C and D Are in Proportion Prove that (13a + 17b)/(13c + 17d) = Sqrt((2ma^2 - 3nb^2)/(2mc^2 - 3nd^2) - Mathematics

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#### Question

If a, b, c and d are in proportion prove that (13a + 17b)/(13c + 17d) = sqrt((2ma^2 - 3nb^2)/(2mc^2 - 3nd^2)

#### Solution

a, b, c and d are in proportion

a/b = c/d = k(say)

Then a = bk and c = dk

L.H.S =  (13a + 17b)/(13c + 17d) = (13(bk) + 17b)/(13(dk) + 17d) = (b(13k + 17))/(d(13k + 17)) = b/d

R.H.S = sqrt((2ma^2 - 3nb^2)/(2mc^2 - 3nd^2)) = sqrt((2m(bk)^2 - 3nb^2)/(2m(dk)^2 - 3nd^2)) = sqrt((b^2(2mk^2 - 3n))/(d^2(2mk^2 - 3n))) = b/d

Hence L.H.S = R.H.S

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