If the roots of the equation $x^3+a x^2+b x+c=0$ are in arithmetic progression, then

If the roots of the equation $x^3+a x^2+b x+c=0$ are in arithmetic progression, then
  1. $a^3-3 a b+c=0$
  2. $9 a b=2 a^3+27 c$
  3. $a^2-2 b c+c=0$
  4. $3 a b-3 c-a^3=0$

Solution

Let the terms of AP be $\mathrm{A}-d, \mathrm{~A}, \mathrm{~A}+d$ $\begin{aligned} & 3 \mathrm{~A}=-a \Rightarrow \mathrm{~A}=\frac{-a}{3} \\ & \mathrm{~A}(\mathrm{~A}-d)+\mathrm{A}(\mathrm{A}+d)+\mathrm{A}^2-d^2=b\end{aligned}$ $3 \mathrm{~A}^2-d^2=b$ $\qquad ...\mathrm{(i)}$ $\mathrm{A}\left(\mathrm{A}^2-d^2\right)=c$ $\Rightarrow \frac{-a}{3}\left(\mathrm{~A}^2+d^2\right)=-c \Rightarrow \mathrm{~A}^2-d^2=\frac{3 c}{a}$ $\qquad ...\mathrm{(ii)}$ From equation (i) - (ii), $\begin{aligned} & 2 \mathrm{~A}^2=b-\frac{3 c}{a} \\ & \frac{2 a^2}{9}=b-\frac{3 c}{a} \Rightarrow 2 a^3=9 a b-27 c \\ & 9 a b=2 a^3+27 c \end{aligned}$

Asked in: AP EAMCET 2024 (20 May Shift 2)

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