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Question
If a, b and c are in continued proportion, prove that `(a^2 + b^2 + c^2)/(a + b + c)^2 = (a - b + c)/(a + b + c)`
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Solution
Given, a, b and c are in continued proportion.
Therefore,
`a/b = b/c`
ac = b2
L.H.S = `(a^2 + b^2 + c^2)/(a + b + c)^2`
= `(a^2 + b^2 + c^2 + 2ab + 2bc + 2ac - (2ab + 2bc + 2ac))/((a + b + c)^2)`
= `((a + b + c)^2 - 2(ab + bc + b^2))/((a + b + c)^2)`
= `((a + b + c)^2 - 2b(a + c + b))/((a + b + c)^2)`
= `((a + b + c)(a + b + c - 2b))/((a + b + c)^2)`
= `(a - b + c)/(a + b + c)`
Hence proved.
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