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$If a x^{3} + b x^{2} + c x + d is divisible by a x^{2} + c, then a, b, c, d \underline{are} in$

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Correct option is D

Since ax3+bx2+cx+d is divisible by ax2+c, therefore, when ax3+bx2+cx+d is divided by ax2+c the remainder should be zero, Now when ax3+bx2+cx+d is divided by ax2+c, then the remainder is bca−d∴bca−d=0⇒bc=ad⇒ba=dc Hence, from this a,b,c,d are not necessarily in G.P.

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