Question

If x, y, z are in continued proportion then show that $\frac{x-y}{x-z}=\frac{y}{y+z}$.

Answer 1

$$\begin{gathered} G\mathrm{iven}: \\ \frac{x}{y}=\frac{y}{z} \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{x-y}{x-z} \\ =\frac{{\displaystyle \frac{x}{y}}-1}{{\displaystyle \frac{x}{y}}-{\displaystyle \frac{z}{y}}}\left(\mathrm{Dividing}\mathrm{by}y\right) \\ =\frac{{\displaystyle \frac{y}{z}}-1}{{\displaystyle \frac{y}{z}}-{\displaystyle \frac{z}{y}}}\left(\because \frac{x}{y}=\frac{y}{z}\right) \\ =\frac{{\displaystyle \frac{y-z}{z}}}{{\displaystyle \frac{y^2-z^2}{zy}}} \\ =\frac{y-z}{z}\times \frac{zy}{\left(y-z\right)\left(y+z\right)} \\ =\frac{y}{y+z} \\ =R.H.S. \\ \therefore \frac{x-y}{x-z}=\frac{y}{y+z} \end{gathered}$$