Question

If $a,b,c$ are in geometric progression, then prove that
$\frac{a}{a+2b}=\frac{a-2b}{a-4c}$

Answer 1

$a,b,c$ are in geometric progression

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

$\Rightarrow b^2=ac$ or $b=\sqrt{ac}$

$\frac{a}{a+2b}=\frac{a}{a+2\sqrt{ac}}=\frac{\sqrt{a}\sqrt{a}}{\sqrt{a}(\sqrt{a}+2\sqrt{c})}$

$=\frac{\sqrt{a}}{\sqrt{a}+2\sqrt{c}}\times \frac{\sqrt{a}-2\sqrt{c}}{\sqrt{a}-2\sqrt{c}}$

$=\frac{\sqrt{a}(\sqrt{a}-2\sqrt{c})}{a-4c}$

$=\frac{a-2\sqrt{ac}}{a-4c}$

$=\frac{a-2\sqrt{b^2}}{a-4c}$ since $ac=b^2$

$=\frac{a-2b}{a-4c}$

Hence proved.