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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
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Verified Answer
The correct answer is:
Arithmetic Progression
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.
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.
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