Solve this


If ${ }^{n} C_{r-1}=36,{ }^{n} C_{r}=84$ and ${ }^{n} C_{r+1}=126$, find $r .$



Given, ${ }^{n} C_{r-1}=36,{ }^{n} C_{r}=84$ and ${ }^{n} C_{r+1}=126$

To find:r=?

$\frac{\left(\begin{array}{c}n \\ r\end{array}\right)}{\left(\begin{array}{c}n \\ r-1\end{array}\right)}=\frac{84}{36}=\frac{7}{3}$

$\frac{\frac{n !}{(n-r) ! \times r !}}{\frac{n !}{(n-r+1) ! \times(r-1) !}}=\frac{7}{3}$

$\Rightarrow \frac{(n-r+1)}{r}=\frac{7}{3}$

$\Rightarrow 3(n-r+1)=7 r$

$\Rightarrow 3 n-10 r=-3 \ldots$(1)

$\Rightarrow \frac{\left(\begin{array}{c}n \\ r+1\end{array}\right)}{\left(\begin{array}{l}n \\ r\end{array}\right)}=\frac{126}{84}=\frac{3}{2}$

$\Rightarrow \frac{\frac{n !}{(n-r-1) ! \times(r+1) !}}{\frac{n !}{(n-r) ! \times r !}}=\frac{3}{2}$

$\Rightarrow \frac{(n-r)}{(r+1)}=\frac{3}{2}$

$\Rightarrow 2(n-r)=3(r+1)$

$\Rightarrow 2 n-5 r=3 \ldots(2)$

From equations 1 & 2 we get

$n=9 \& r=3$

Ans: $r=3$


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