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If the sum of the roots of the quadratic equation $a x^2+b x+c=0$ is equal to the sum of the squares of their reciprocals, then $\frac{\mathrm{a}}{\mathrm{c}}, \frac{\mathrm{b}}{\mathrm{a}}$ and $\frac{\mathrm{c}}{\mathrm{b}}$ are in
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Harmonic Progression
Harmonic Progression
$a x^2+b x+c=0, \alpha+\beta=\frac{-b}{a}, \alpha \beta=\frac{c}{a}$
As for given condition, $\alpha+\beta=\frac{1}{\alpha^2}+\frac{1}{\beta^2}$
$\alpha+\beta=-\frac{\alpha^2+\beta^2}{\alpha^2 \beta^2}-\frac{b}{a}=\frac{\frac{b^2}{a^2}-\frac{2 c}{a}}{\frac{c^2}{a^2}}$
On simplification $2 a^2 c=a b^2+b c^2$
$\Rightarrow \frac{2 \mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{c}} \quad \therefore \frac{\mathrm{a}}{\mathrm{b}}, \frac{\mathrm{b}}{\mathrm{a}}, \& \frac{\mathrm{c}}{\mathrm{b}}$ are in H.P.
As for given condition, $\alpha+\beta=\frac{1}{\alpha^2}+\frac{1}{\beta^2}$
$\alpha+\beta=-\frac{\alpha^2+\beta^2}{\alpha^2 \beta^2}-\frac{b}{a}=\frac{\frac{b^2}{a^2}-\frac{2 c}{a}}{\frac{c^2}{a^2}}$
On simplification $2 a^2 c=a b^2+b c^2$
$\Rightarrow \frac{2 \mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{c}} \quad \therefore \frac{\mathrm{a}}{\mathrm{b}}, \frac{\mathrm{b}}{\mathrm{a}}, \& \frac{\mathrm{c}}{\mathrm{b}}$ are in H.P.
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