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If the coefficients of $2^{nd},3^{rd}\:and\:4^{th}$ terms in the expansion of $(1+x)^{2n} $ are in A.P., then $n=?$

$\begin{array}{1 1} 1 \\ \frac{5}{2} \\ \frac{7}{2} \\ No\;such\;n\;exists \end{array} $

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Given: $^{2n}C_1,^{2n}C_2, ^{2n}C_3$ are in A.P.
$\Rightarrow\:2.^{2n}C_2=^{2n}C_1+^{2n}C_3$
$\Rightarrow\:2.\large\frac{(2n)!}{(2n-2)!.2!}=\frac{(2n)!}{(2n-1)!}+\frac{(2n)!}{(2n-3)!.3!}$
$\Rightarrow\:\large\frac{1}{2n-2}=\frac{6+(2n-1)(2n-2)}{3!.(2n-1)(2n-2)}$
$\Rightarrow\:12n-6=6+4n^2-6n+2$
$\Rightarrow\:4n^2-18n+14=0$
$\Rightarrow\:2n^2-9n+7=0$
$\Rightarrow\:n=1\:or\:n=\large\frac{7}{2}$
But since $n$ cannot be 1 $n=\large\frac{7}{2}$
answered Oct 1, 2013 by rvidyagovindarajan_1
 

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