If three distinct numbers a, b, c are

Since $a, b, c$ are in G.P.

$\therefore b^{2}=a c$

Given equation is, $a x^{2}+2 b x+c=0$

$\Rightarrow a x^{2}+2 \sqrt{a c} x+c=0 \Rightarrow(\sqrt{a} x+\sqrt{c})^{2}=0$

$\Rightarrow x=-\sqrt{\frac{c}{a}}$

Also, given that $a x^{2}+2 b x+c=0$ and $d x^{2}+2 e x+f=0$ have a common root.

$\Rightarrow x=-\sqrt{\frac{c}{a}}$ must satisfy $d x^{2}+2 e x+f=0$

$\Rightarrow d \cdot \frac{c}{a}+2 e\left(-\sqrt{\frac{c}{a}}\right)+f=0$

$\frac{d}{a}-\frac{2 e}{\sqrt{a c}}+\frac{f}{c}=0 \quad \Rightarrow \frac{d}{a}-\frac{2 e}{b}+\frac{f}{c}=0$

$\Rightarrow \frac{2 e}{b}=\frac{d}{a}+\frac{f}{c} \Rightarrow \frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.