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If $a, b, c$ are distinct and the roots of $(b-c) x^2+(c-a) x$ $+(a-b)=0$ are equal, then $a, b$ and $c$ are in
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The correct answer is:
arithmetic progression
Given that the roots of equation $(b-c) x^2+(c-a) x+$ $(a-b)=0$ are equal, so
$\begin{aligned}
& D=0 \\
& (c-a)^2-4(a-b)(b-c)=0 \\
& c^2+a^2-2 c a-4 a b+4 a c+4 b^2=0 \\
& c^2+a^2+2 a c+4 b^2-4 b(c+a)=0 \\
& (c+a)^2+(2 b)^2-2 \cdot 2 b(c+a)=0 \\
& {[(c+a)-(2 b)]^2=0} \\
& c+a-2 b=0 \\
& 2 b=a+c
\end{aligned}$
Hence, we can conclude that $a, b$ and $c$ are in AP.
$\begin{aligned}
& D=0 \\
& (c-a)^2-4(a-b)(b-c)=0 \\
& c^2+a^2-2 c a-4 a b+4 a c+4 b^2=0 \\
& c^2+a^2+2 a c+4 b^2-4 b(c+a)=0 \\
& (c+a)^2+(2 b)^2-2 \cdot 2 b(c+a)=0 \\
& {[(c+a)-(2 b)]^2=0} \\
& c+a-2 b=0 \\
& 2 b=a+c
\end{aligned}$
Hence, we can conclude that $a, b$ and $c$ are in AP.
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