Question

If $a,b,c$ are in continued proportion prove that:
(i) $\frac{a^2+ab+b^2}{b^2+bc+c^2}=\frac{a}{c}$
(ii) $\frac{a^2+b^2+c^2}{{(a+b+c)}^2}=\frac{a-b+c}{a+b+c}$

Answer 1

Let ratio of proportion be $r$

$⟹\frac{b}{a}=\frac{c}{b}=r$

$⟹b=ar$ and $c=br=a{r}^2$

$\left(i\right)$ $LHS=\frac{a^2+ab+b^2}{b^2+bc+c^2}=\frac{a^2+a^{2}r+a^2{r}^2}{a^2{r}^2+a^2{r}^3+a^2{r}^4}=\frac{1}{r^2}$

$RHS=\frac{a}{c}=\frac{a}{a{r}^2}=\frac{1}{r^2}$

Hence $LHS=RHS$

$\left(ii\right)$ $LHS=\frac{a^2+b^2+c^2}{(a+b+c{)}^2}=\frac{a^2+a^2{r}^2+a^2{r}^4}{(a+ar+a{r}^2{)}^2}=\frac{1+r^2+r^4}{(1+r+r^2{)}^2}$

$RHS=\frac{a-b+c}{a+b+c}=\frac{a-ar+a{r}^2}{a+ar+a{r}^2}=\frac{1-r+r^2}{1+r+r^2}$

$=\frac{(1+r^2)-r}{1+r+r^2}\times \frac{(1+r^2)+r}{1+r+r^2}=\frac{(1+r^2{)}^2-r^2}{(1+r+r^2{)}^2}$

$=\frac{1+r^2+r^4}{(1+r+r^2{)}^2}$

Hence $LHS=RHS$