Question

If $x,y,z$ are three consecutive terms in a G.P prove that
$\frac{1}{x+y}+\frac{1}{y+z}=\frac{1}{y}$

Answer 1

Let x,y,z are in G.P

and x,y,z are 3 consecutive terms.

Let r be the common difference.

$\Rightarrow y=xr$ and $z=x{r}^2$

Now $\frac{1}{x+y}+\frac{1}{y+z}$

$=\frac{1}{x+xr}+\frac{1}{xr+x{r}^2}$

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

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

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

$=\frac{1}{xr}=\frac{1}{y}$

$\therefore \frac{1}{xy}+\frac{1}{y+z}=\frac{1}{y}$ (proved)