Condition for the roots of the equation $x^{3} - p x^{2} + q x - r = 0$ are in $G . P$ is

q = p r

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q^{2} = p^{2} r

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

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

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Solution

The correct option is D $q^{3} = p^{3} r$
Let $α, β, γ$ be roots of $v$
such that $β^{2} = α γ$
Then
$s_{3} = α β γ = r \Rightarrow β^{3} = r \Rightarrow β = \sqrt[3]{r}$
Substituting this in equation we get
${(\sqrt[3]{r})}^{3} - p {(\sqrt[3]{r})}^{2} + q (\sqrt[3]{r}) - r = 0 \Rightarrow r - p r^{\frac{2}{3}} + q r^{\frac{1}{3}} - r = 0 \Rightarrow q^{3} = p^{3} r$

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