If a, b, c are in GP, prove that

Solution:

To prove: $\frac{a^{2}+a b+b^{2}}{a b+b c+c a}=\frac{b+a}{c+b}$

Given: a, b, c are in GP

Formula used: When $a, b, c$ are in GP, $b^{2}=a c$

a, b, c are in GP

$\Rightarrow \mathrm{b}^{2}=\mathrm{ac} \ldots$ (i)

$\Rightarrow \mathrm{b}=\sqrt{\mathrm{ac}}$ … (ii)

Taking LHS = $\frac{a^{2}+a b+b^{2}}{a b+b c+c a}$

Substituting the value $b^{2}=a c$ from eqn. (i)

$\mathrm{LHS}=\frac{\mathrm{a}^{2}+\mathrm{ab}+\mathrm{ac}}{\mathrm{ab}+\mathrm{bc}+\mathrm{b}^{2}}$

$\Rightarrow \frac{a(a+b+c)}{b(a+b+c)}$

$\Rightarrow \frac{a}{b}$

Substituting the value $b=\sqrt{a c}$ from eqn. (ii)

$\Rightarrow \frac{a}{\sqrt{a c}}$

$\Rightarrow \frac{\sqrt{a}}{\sqrt{c}}$

Multiplying and dividing with $(\sqrt{a}+\sqrt{c})$

$\Rightarrow \frac{\sqrt{a}(\sqrt{a}+\sqrt{c})}{\sqrt{c}(\sqrt{a}+\sqrt{c})}$

$\Rightarrow \frac{(a+\sqrt{a c})}{(\sqrt{a c}+c)}$

$\Rightarrow \frac{a+b}{b+c}=R H S$

Hence Proved