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Q.

The condition that the roots of $x^{3} + 3 p x^{2} + 3 q x + r = 0$ may be in G.P is

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a

$p^{3} r = q^{3}$

b

$p^{3} r = q$

c

$p r = q^{3}$

d

$p^{3} r^{3} = q^{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{a}{k}, a, a k \\ \frac{a}{k} . a . a k = - \frac{d}{a} \\ a^{3} = - k \\ a = {(- k)}^{\frac{1}{3}} \\ f (a) = 0 \\ {(\sqrt[3]{- k})}^{3} + 3 p {(\sqrt[3]{- k})}^{2} + 3 q (\sqrt[3]{- k}) + r = 0 \\ p^{3} r^{2} = q^{3} r \\ p^{3} r = q^{3} \end{array}$

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