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If the coefficients of $r$ th, $(r+1)$ th and $(r+2)$ th terms in the binomial expansion of $(1+$ $y)^m$ are in A.P., then $m$ and $r$ satisfy the equation
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The correct answer is:
$m^2-m(4 r+1)+4 r^2-2=0$
$m^2-m(4 r+1)+4 r^2-2=0$
Given ${ }^m C,{ }^m C,{ }^m C_{r+1}$ are in A.P.
$$
\begin{aligned}
& 2{ }^m C_r={ }^m C_{r-1}+{ }^m C_{r+1} \\
& \Rightarrow 2=\frac{{ }^m C_{r-1}}{{ }^m C_r}+\frac{{ }^m C_{r+1}}{{ }^m C_r} \\
& =\frac{r}{m-r+1}+\frac{m-r}{r+1} \\
& \Rightarrow m^2-m(4 r+1)+4 r^2-2=0
\end{aligned}
$$
$$
\begin{aligned}
& 2{ }^m C_r={ }^m C_{r-1}+{ }^m C_{r+1} \\
& \Rightarrow 2=\frac{{ }^m C_{r-1}}{{ }^m C_r}+\frac{{ }^m C_{r+1}}{{ }^m C_r} \\
& =\frac{r}{m-r+1}+\frac{m-r}{r+1} \\
& \Rightarrow m^2-m(4 r+1)+4 r^2-2=0
\end{aligned}
$$
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