Question

If a, b, c are in G.P. then prove that : $\frac{a^2+ab+b^2}{bc+ca+ab}=\frac{b+a}{c+b}$

Answer 1

Since, a, b, c are in G.P.

$\therefore$ a = a, b = ar, $c=a{r}^2$

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

$\frac{a^2+a\left(ar\right)+a^2{r}^2}{\left(ar\right)(a{r}^{)}+(a{r}^2)+a+(ar)}=\frac{ar+a}{a{r}^2+ar}$

$\frac{a^2+(1+r+r^2)}{a^2(r^3+r^2+r)}=\frac{a(1+r)}{a(r^2+r)}$

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

$\frac{1}{r}=\frac{1}{r}$

LHS = RHS

Hence,

$\frac{a^2+ab+b^2}{bc+ca+ab}=\frac{b+a}{c+d}$