$If p_{1}, p_{2} are the roots of the quadratic equation a x^{2} + b x + c = 0 and q_{1}, q_{2} are the$ $roots of quadratic equation c x^{2} + b x + a = 0 (a, b, c \in R) . Such that p_{1}, q_{1}, p_{2}, q_{2} are$
$in A.P of distinct terms then (a / c) equals:$

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-1

$\frac{1}{2}$

answer is B.

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Detailed Solution

$p_{1}, p_{2} are roots of a x^{2} + b x + c = 0$

$q_{1}, q_{2} are roots of c x^{2} + b x + a = 0$

$Put y = \frac{1}{x}, so c + b y + a y^{2} = 0$

$Put \begin{array}{l} y = \frac{1}{x}, so c + b y + a y^{2} = 0 \\ q_{1} = \frac{1}{p_{1}} and q_{2} = \frac{1}{p_{2}} \\ p_{1} p_{2} = \frac{c}{a}; p_{1} + p_{2} = - \frac{b}{a} \\ q_{1} q_{2} = \frac{a}{c}; q_{1} + q_{2} = - \frac{b}{c} \\ p_{1}, q_{1}, p_{2}, q_{2} are in A.P \\ {(p_{2} - p_{1})}^{2} = {(q_{2} - q_{1})}^{2} \\ \frac{b^{2}}{a^{2}} - 4 \frac{c}{a} = \frac{b^{2}}{c^{2}} - \frac{4 a}{c} \\ \frac{b^{2} - 4 a c}{a^{2}} = \frac{b^{2} - 4 a c}{c^{2}} \\ \frac{a^{2}}{c^{2}} = 1 \\ Thus, \frac{a}{c} = - 1 \end{array}$