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(C) Let the roots be $\alpha$ and $\beta$. Then $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$.Given $\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{

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harmonic progression $(H.P.)$

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(C) Let the roots be $\alpha$ and $\beta$. Then $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$.Given $\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$.Substituting the values: $-b/a = \frac{(-b/a)^2 - 2(c/a)}{(c/a)^2} = \frac{b^2/a^2 - 2c/a}{c^2/a^2} = \frac{b^2 - 2ac}{c^2}$.$-b/a = \frac{b^2 - 2ac}{c^2} \Rightarrow -bc^2 = ab^2 - 2a^2c$.$2a^2c = ab^2 + bc^2$.Dividing by $abc$: $\frac{2a}{b} = \frac{b}{c} + \frac{c}{a}$.This implies $\frac{c}{a}, \frac{a}{b}, \frac{b}{c}$ are in $A.P.$Therefore,their reciprocals $\frac{a}{c}, \frac{b}{a}, \frac{c}{b}$ 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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