Question

If $a,bandc$ in continued proportion, prove that:
$\frac{a^2+ab+b^2}{b^2+bc+c^2}=\frac{a}{c}$

Answer 1

prove that:
$\frac{a^2+ab+b^2}{b^2+bc+c^2}=\frac{a}{c}$

If $a,bandc$ in continued proportion then:$\frac{a}{b}=\frac{b}{c}$

let $\frac{a}{b}=\frac{b}{c}=k$

$\frac{a}{b}=k$ and $\frac{b}{c}=k$

$a=bkandb=ck$

$a=\left(ck\right)k=c{k}^2$

Considering LHS=$\frac{a^2+ab+b^2}{b^2+bc+c^2}$

$=\frac{{\left(c{k}^2\right)}^2+c{k}^2\cdot ck+{\left(ck\right)}^2}{{\left(ck\right)}^2+ck\cdot c+c^2}$

$=\frac{c^2{k}^2(k^2+k+1)}{c^2(k^2+k+1)}=k^2$

Considering RHS$=\frac{a}{c}=\frac{c{k}^2}{c}=k^2$

LHS = RHS

Hence, proved.