If a, b, c are in geometric progression,then | vedclass

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(A) Since $a, b, c$ are in geometric progression,we have $b^2 = ac$.The equation $ax^2 + 2bx + c = 0$ becomes $ax^2 + 2\sqrt{ac}x + c = 0$,which is $(

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Arithmetic Progression

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(A) Since $a, b, c$ are in geometric progression,we have $b^2 = ac$.The equation $ax^2 + 2bx + c = 0$ becomes $ax^2 + 2\sqrt{ac}x + c = 0$,which is $(\sqrt{a}x + \sqrt{c})^2 = 0$.Thus,the roots are $x = -\sqrt{\frac{c}{a}}, -\sqrt{\frac{c}{a}}$.For the equation $dx^2 + 2ex + f = 0$ to have the same roots,the root $-\sqrt{\frac{c}{a}}$ must satisfy it.Substituting $x = -\sqrt{\frac{c}{a}}$ into $dx^2 + 2ex + f = 0$,we get $d(\frac{c}{a}) - 2e\sqrt{\frac{c}{a}} + f = 0$.Dividing by $c$,we get $\frac{d}{a} - \frac{2e}{c}\sqrt{\frac{c}{a}} + \frac{f}{c} = 0$.Since $\sqrt{\frac{c}{a}} = \frac{c}{\sqrt{ac}} = \frac{c}{b}$,we have $\frac{d}{a} + \frac{f}{c} = \frac{2e}{c} \cdot \frac{c}{b} = \frac{2e}{b}$.This is the condition for $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ to be in Arithmetic Progression.

If $a, b, c$ are in geometric progression,then in which progression are $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ such that the equations $ax^2 + 2bx + c = 0$ and $dx^2 + 2ex + f = 0$ have common roots?

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