The condition that the roots of $a x^{3} + b x^{2} + c x + d = 0$ may be in A.P. is

2 b^{3} + 27 a^{2} d = 9 a b c

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2 b^{3} + 27 a^{2} d = - 9 a b c

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2 b^{3} - 27 a^{2} d = 9 a b c

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2 b^{3} - 27 a^{2} d = - 9 a b c

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Solution

The correct option is A $2 b^{3} + 27 a^{2} d = 9 a b c$
Let the roots be $p, q, r$ . The roots are in A.P.
So,
$2 q = p + r,$
$3 q = p + q + r .$
Using theory of equations,
$p + q + r = \frac{- b}{a} .$
So, $q = \frac{- b}{3 a} .$
But, $q$ is a root of the given equation. So
$a q^{3} + b q^{2} + c q + d = 0,$
$a {(\frac{- b}{3 a})}^{3} + b {(\frac{- b}{3 a})}^{2} + c (\frac{- b}{3 a}) + d = 0,$
Rearranging, we get
$2 b^{3} + 27 a^{2} d = 9 a b c .$

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