#### Question

If a, b, and c are in continued proportion, prove that `(a^2 + ab + b^2)/(b^2 + bc + c^2) = a/c`

#### Solution

Given, a, b and c are in continued proportion.

=> `a/b = b/c = k` (say)

=> a = bk, b = ck

=> a = `(ck)k = ck^2`, b = ck

L.H.S `(a^2 + ab + b^2)/(b^2 + bc + c^2)`

`= ((ck^2)^2 + (ck^2)(ck) + (ck)^2)/((ck)^2 + (ck)c + c^2)`

`= (c^2k^4 + c^2k^3 + c^2k^2)/(c^2k^2 + c^2k + c^2)`

`= (c^2k^2(k^2 + k + 1))/(c^2(k^2 + k + 1))`

`= k^2`

RHS = `a/c = (ck^2)/c = k^2`

∴ LHS = RHS

Is there an error in this question or solution?

Solution If A, B, and C Are in Continued Proportion, Prove that (A^2 + Ab + B^2)/(B^2 + Bc + C^2) = A/C Concept: Proportions.