Question

$\frac{p+q}{p-q}$, x and p² − q² are in continued proportion then find x.

Answer 1

$$\begin{gathered} \frac{p+q}{p-q},x\mathrm{and}{p}^2-q^2\mathrm{are}\mathrm{in}\mathrm{continued}\mathrm{proportion}. \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ \frac{{\displaystyle \frac{p+q}{p-q}}}{x}=\frac{x}{p^2-q^2} \\ \Rightarrow x^2=\left(\frac{p+q}{p-q}\right)\times \left(p^2-q^2\right) \\ \Rightarrow x^2={\left(p+q\right)}^2 \\ \Rightarrow x=\left(p+q\right) \end{gathered}$$