Question

If a, b, c an in G.P then find the value of $\frac{b-a}{b-c}+\frac{b+a}{b+c}$:

Answer 1

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

$\Rightarrow b^2=ac$

Then $\frac{b-a}{b-c}+\frac{b+a}{b+c}$

$\Rightarrow \frac{b^2+bc-ab-ac+b^2+ab-bc-ac}{b^2-c^2}$

$\Rightarrow \frac{2b^2-2ac}{b^2-c^2}$

$\Rightarrow \frac{2ac-2ac}{b^2-c^2}=0$ $\because b^2=ac$

$\therefore \frac{b-a}{b-c}+\frac{b+a}{b+c}=0$