If three distinct numbers a, b, c are in G.P. | vedclass

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(A) Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.The equation $ax^2 + 2bx + c = 0$ can be written as $ax^2 + 2\sqrt{ac}x + c = 0$,which is $(\sqrt

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

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(A) Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.The equation $ax^2 + 2bx + c = 0$ can be written as $ax^2 + 2\sqrt{ac}x + c = 0$,which is $(\sqrt{a}x + \sqrt{c})^2 = 0$.Thus,the root of this equation is $x = -\frac{\sqrt{c}}{\sqrt{a}} = -\frac{b}{a}$.Since this is a common root for $dx^2 + 2ex + f = 0$,we substitute $x = -\frac{b}{a}$ into the second equation:$d(-\frac{b}{a})^2 + 2e(-\frac{b}{a}) + f = 0$$d(\frac{b^2}{a^2}) - \frac{2eb}{a} + f = 0$Multiplying by $a^2$,we get $db^2 - 2eab + fa^2 = 0$.Since $b^2 = ac$,we substitute it:$dac - 2eab + fa^2 = 0$.Dividing the entire equation by $ac$,we get:$\frac{d}{a} - \frac{2e}{b} + \frac{f}{c} = 0$,which implies $\frac{d}{a} + \frac{f}{c} = 2(\frac{e}{b})$.This condition indicates that $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in $A.P.$

If three distinct numbers $a, b, c$ are in $G.P.$ and the equations $ax^2 + 2bx + c = 0$ and $dx^2 + 2ex + f = 0$ have a common root,then which one of the following statements is correct?

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