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)`

Answer 1

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.