If the roots of the equation $6 x^3-11 x^2+6 x-1=0$ are in harmonic progression, then the roots of $x^3-6…

If the roots of the equation $6 x^3-11 x^2+6 x-1=0$ are in harmonic progression, then the roots of $x^3-6 x^2+11 x-$ $6=0$ will be in
  1. Geometric Progression
  2. Arithmetic Progression
  3. Harmonic Progression
  4. Arithmetico-Geometric Progression

Solution

Consider $x^3-6 x^2+11 x-6=0$ ...(i) Since $x=1$, satisfies the above equation. Hence $(x-1)$ will be a factor of equation (i) Therefore $ \begin{aligned} & x^2(x-1)-5 x(x-1)+6(x-1)=0 \\ & \Rightarrow\left(x^2-5 x+6\right)(x-1)=0 \\ & \Rightarrow(x-3)(x-2)(x-1)=0 \Rightarrow x=1,2,3 \end{aligned} $ Since $2-1=1=3-2$ Hence roots of equation is in Arithmetic progression

Asked in: AP EAMCET 2023 (19 May Shift 1)

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