If $p, q, r$ are in $G.P.$ and the equations $p x^{2}+2 q x+r=0$ and $d x^{2}+2 e x+f=0$ have a common root,then show that $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in $A.P.$

(N/A) Given that $p, q, r$ are in $G.P.$,we have $q^{2} = pr$.
The equation $p x^{2} + 2 q x + r = 0$ can be written as $p x^{2} + 2 q x + \frac{q^{2}}{p} = 0$,which simplifies to $p^{2} x^{2} + 2 p q x + q^{2} = 0$.
This is $(p x + q)^{2} = 0$,so the root is $x = -\frac{q}{p}$.
Since this is a common root for $d x^{2} + 2 e x + f = 0$,we substitute $x = -\frac{q}{p}$ into the second equation:
$d(-\frac{q}{p})^{2} + 2 e(-\frac{q}{p}) + f = 0$
$d(\frac{q^{2}}{p^{2}}) - 2 e(\frac{q}{p}) + f = 0$
Multiply by $\frac{p^{2}}{q^{2}}$:
$d - 2 e(\frac{p}{q}) + f(\frac{p^{2}}{q^{2}}) = 0$
Since $q^{2} = pr$,we have $\frac{p}{q} = \frac{q}{r}$,so $\frac{p^{2}}{q^{2}} = \frac{p}{r}$.
$d - 2 e(\frac{p}{q}) + f(\frac{p}{r}) = 0$
Divide by $p$:
$\frac{d}{p} - 2 \frac{e}{q} + \frac{f}{r} = 0$
$\frac{d}{p} + \frac{f}{r} = 2 \frac{e}{q}$
This confirms that $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in $A.P.$