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 an A.P.

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Solution

$p, q, r \to G . P$
$\Rightarrow q^{2} = p r$ .... (1)
$p x^{2} + 2 q x + r = 0$
$d x^{2} + 2 e x + f = 0$ common roots
Condn of common root : -
$(r d - P f)^{2} = (4 p e - 4 q d) (4 q f - 4 e r)$
$\Rightarrow r^{2} d^{2} + p^{2} f^{2} - 2 r d p f = 16 [p e q f - p r e^{2} - q^{2} d f - q d e r]$
$\Rightarrow {(\frac{2 e}{q} - \frac{d}{p} - \frac{f}{r})}^{2} = 0$
$\Rightarrow \frac{2 e}{q} = \frac{d}{p} + \frac{f}{r}$
Hence proved.

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