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(C) Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$. Then $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$.Given th

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harmonic progression $(H.P.)$

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(C) Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$. Then $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$.Given that $\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2}$.Substituting the values,we get $-b/a = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} = \frac{(-b/a)^2 - 2(c/a)}{(c/a)^2} = \frac{b^2/a^2 - 2c/a}{c^2/a^2} = \frac{b^2 - 2ac}{c^2}$.Thus,$-b/a = \frac{b^2 - 2ac}{c^2} \Rightarrow -bc^2 = ab^2 - 2a^2c$.Rearranging,$2a^2c = ab^2 + bc^2$. Dividing both sides by $abc$,we get $2a/b = b/c + c/a$.This implies $c/a, a/b, b/c$ are in $A$.$P$. Therefore,their reciprocals $a/c, b/a, c/b$ are in $H$.$P$.

If the sum of the roots of the quadratic equation $ax^2 + bx + c = 0, (abc \neq 0)$ is equal to the sum of the squares of their reciprocals,then $a/c, b/a, c/b$ are in

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