If q is the mean proportional between p and r, prove that :

$p^2 - 3q^2 + r^2 = q^4\Big(\dfrac{1}{p^2} - \dfrac{3}{q^2} + \dfrac{1}{r^2}\Big).$

Since, q is the mean proportional between p and r,

$\therefore q^2 = pr$

Given,

$\Rightarrow p^2 - 3q^2 + r^2 = q^4\Big(\dfrac{1}{p^2} - \dfrac{3}{q^2} + \dfrac{1}{r^2}\Big). \\[1em] \text{L.H.S.} = p^2 - 3q^2 + r^2 \\[1em] = p^2 + r^2 - 3pr. \\[1em] \text{R.H.S.} = q^4\Big(\dfrac{1}{p^2} - \dfrac{3}{q^2} + \dfrac{1}{r^2}\Big) \\[1em] \Rightarrow p^2r^2\Big(\dfrac{1}{p^2} - \dfrac{3}{pr} + \dfrac{1}{r^2}\Big) \\[1em] \Rightarrow p^2r^2\Big(\dfrac{r^2 - 3pr + p^2}{p^2r^2}\Big) \\[1em] \Rightarrow p^2 + r^2 - 3pr.$

Since, L.H.S. = R.H.S. hence proved that,

$p^2 - 3q^2 + r^2 = q^4\big(\dfrac{1}{p^2} - \dfrac{3}{q^2} + \dfrac{1}{r^2}\big).$