If the roots of x^3-p x^2+q x-r=0 are in AP,t | vedclass

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(A) Let the roots be $a-d, a, a+d$. Since the roots are in $AP$,their sum is given by the coefficient of $x^2$: $(a-d) + a + (a+d) = p$ $3a = p \impli

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$2 p^3-9 p q+27 r=0$

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(A) Let the roots be $a-d, a, a+d$. Since the roots are in $AP$,their sum is given by the coefficient of $x^2$: $(a-d) + a + (a+d) = p$ $3a = p \implies a = \frac{p}{3}$. Since $a$ is a root of the equation $x^3-p x^2+q x-r=0$,it must satisfy the equation: $(\frac{p}{3})^3 - p(\frac{p}{3})^2 + q(\frac{p}{3}) - r = 0$ $\frac{p^3}{27} - \frac{p^3}{9} + \frac{pq}{3} - r = 0$ Multiply the entire equation by $27$: $p^3 - 3p^3 + 9pq - 27r = 0$ $-2p^3 + 9pq - 27r = 0$ $2p^3 - 9pq + 27r = 0$. Thus,the correct option is $A$.

If the roots of $x^3-p x^2+q x-r=0$ are in $AP$,then:

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