If a, b and c are in continued proportion, prove that :

$\dfrac{a^2 + ab + b^2}{b^2 + bc + c^2} = \dfrac{a}{c}$

Given,

a, b and c are in continued proportion

$\therefore \dfrac{a}{b} = \dfrac{b}{c} \\[1em] \text{Let } \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow a = bk, b = ck \\[1em] \Rightarrow a = (ck)k = ck^2$

To prove :

$\dfrac{a^2 + ab + b^2}{b^2 + bc + c^2} = \dfrac{a}{c}$

Substituting value of a and b in L.H.S. of above equation,

$\Rightarrow \dfrac{(ck^2)^2 + (ck^2)(ck) + (ck)^2}{(ck)^2 + (ck)c + c^2} \\[1em] \Rightarrow \dfrac{c^2k^4 + c^2k^3 + c^2k^2}{c^2k^2 + c^2k + c^2} \\[1em] \Rightarrow \dfrac{c^2k^2(k^2 + k + 1)}{c^2(k^2 + k + 1)} \\[1em] \Rightarrow k^2.$

Substituting value of a and b in R.H.S. of above equation,

$\Rightarrow \dfrac{ck^2}{c} \\[1em] \Rightarrow k^2.$

Since, L.H.S. = R.H.S. = k².

Hence, proved that $\dfrac{a^2 + ab + b^2}{b^2 + bc + c^2} = \dfrac{a}{c}$.