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

$$\begin{gathered} a,b\mathrm{and}c\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}. \\ \therefore b^2=ac........\left(\mathrm{i}\right) \\ \mathrm{Now},\mathrm{LHS}=\frac{{\mathrm{a}}^2+\mathrm{ab}+{\mathrm{b}}^2}{\mathrm{bc}+\mathrm{ca}+\mathrm{ab}} \\ =\frac{{\mathrm{a}}^2+\mathrm{ab}+\mathrm{ac}}{\mathrm{bc}+{\mathrm{b}}^2+\mathrm{ab}}\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ =\frac{a\left(a+b+c\right)}{b\left(c+b+a\right)} \\ =\frac{a}{b} \\ =\frac{1}{r} \\ \mathrm{Here},r=\mathrm{common}\mathrm{ratio} \\ \mathrm{RHS}=\frac{b+a}{c+b} \\ =\frac{ar+a}{a{r}^2+ar} \\ =\frac{a(r+1)}{ar(r+1)} \\ =\frac{1}{r} \\ \therefore \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$