If in the equation ax^2 + bx + c = 0,the sum | vedclass

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

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$A.P.$

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(A) Let the roots be $\alpha$ and $\beta$. Then $\alpha + \beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$.Given that $\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}$.$\alpha + \beta = \frac{\alpha^2 + \beta^2}{(\alpha \beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha \beta}{(\alpha \beta)^2}$.Substituting the values of $\alpha + \beta$ and $\alpha \beta$:$-\frac{b}{a} = \frac{(-b/a)^2 - 2(c/a)}{(c/a)^2} = \frac{b^2/a^2 - 2c/a}{c^2/a^2} = \frac{b^2 - 2ac}{c^2}$.$-bc^2 = a(b^2 - 2ac) = ab^2 - 2a^2c$.$2a^2c = ab^2 + bc^2$.Dividing both sides by $abc$:$\frac{2a^2c}{abc} = \frac{ab^2}{abc} + \frac{bc^2}{abc}$.$\frac{2a}{b} = \frac{b}{c} + \frac{c}{a}$.This implies that $\frac{c}{a}, \frac{a}{b}, \frac{b}{c}$ are in $A.P.$ since $2(	ext{middle term}) = 	ext{first term} + 	ext{third term}$.

If in the equation $ax^2 + bx + c = 0$,the sum of roots is equal to the sum of the squares of their reciprocals,then $\frac{c}{a}, \frac{a}{b}, \frac{b}{c}$ are in

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