Given, a, b, c are in G.P.

$$ \therefore $$ b² = ac

In this equation ax² + 2bx + c = 0,

Discrimant, D = 4b² - 4ac

= 4ac - 4ac

= 0

Discrimant = 0 meand roots of the equation are equal.

Let both the roots of the equation = $$\alpha $$

$$ \therefore $$ 2$$\alpha $$ = $$ - {{2b} \over a}$$

$$ \Rightarrow $$ $$\alpha $$ = $$ - {b \over a}$$

As both the equations ax² + 2bx + c = 0 and dx² + 2ex + ƒ = 0 have a common root,

so $$ - {b \over a}$$ is also root of the equation dx² + 2ex + ƒ = 0.

$$ \therefore $$ $$ - {b \over a}$$ satisfy the equation dx² + 2ex + ƒ = 0.

$$ \therefore $$ $$d{\left( { - {b \over a}} \right)^2} + 2e\left( { - {b \over a}} \right) + f = 0$$

$$ \Rightarrow $$ $$d{b^2} - 2aeb + f{a^2} = 0$$

$$ \Rightarrow $$ $$dac - 2aeb + f{a^2} = 0$$

$$ \Rightarrow $$ $$dc - 2eb + fa = 0$$

$$ \Rightarrow $$ $${{dc} \over {ac}} - {{2eb} \over {ac}} + {{fa} \over {ac}} = 0$$

$$ \Rightarrow $$ $${{dc} \over {ac}} - {{2eb} \over {{b^2}}} + {{fa} \over {ac}} = 0$$

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

$$ \Rightarrow $$ $${{2e} \over b} = {d \over a} + {f \over c}$$

$$ \therefore $$ $$d \over a$$, $$e \over b$$, $$f \over c$$ are in A.P.