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If $\large\frac{T_2}{T_3}$ in the expansion of $(a+b)^n$ and $\large\frac{T_3}{T_4}$ in the expansion of $(a+b)^{n+3}$ are equal, then $n=?$

$\begin{array}{1 1} 3 \\ 4 \\ 5 \\ 6 \end{array} $

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$T_2$ in the expansion of $(a+b)^n$ is $^nC_1.a^{n-1}.b$ and
$T_3=^nC_2.a^{n-2}.b^2$
$\large\frac{T_2}{T_3}=\frac{2a}{(n-1)b}$...............(i)
$T_3$ in the expansion of $(a+b)^{n+3}$ is $^{n+3}C_2.a^{n+1}.b^2$ and
$T_4=^{n+3}C_3.a^{n}.b^3$
$\large\frac{T_3}{T_4}=\frac{3a}{(n+1)b}$............(ii)
Given: (i) = (ii)
$\Rightarrow\:\large\frac{2a}{(n-1)b}=\frac{3a}{(n+1)b}$
$\Rightarrow\:2(n+1)=3(n-1)$
$\Rightarrow\:n=5$
answered Sep 26, 2013 by rvidyagovindarajan_1
 

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