Question

If x, y, z are in continued proportion, prove that : $\frac{(x+y{)}^2}{(y+z{)}^2}=\frac{x}{z}$

Answer 1

$x,y,z$ are in continued proportion, that is

$\frac{x}{y}=\frac{y}{z}$ .......... (1)

then,$y^2=xz$

Now,$\frac{(x+y{)}^2}{(y+z{)}^2}$ $=\frac{y^2{(\frac{x}{y}+1)}^2}{z^2{(\frac{y}{z}+1)}^2}$

$=\frac{y^2{(\frac{x}{y}+1)}^2}{z^2{(\frac{x}{y}+1)}^2}$ ( from (1) $\frac{x}{y}=\frac{y}{z}$ )

$=\frac{y^2}{z^2}$

$=\frac{xz}{z^2}$

$=\frac{x}{z}$