JEE Main 2019 — Binomial Theorem Question with Solution

From the given condition

${ }^n{C}_r{:}^n{C}_{r+1}{:}^n{C}_{r+2}=2:15:70$

$\Rightarrow \frac{{ }^n{C}_r}{{ }^n{C}_{r+1}}=\frac{2}{15}$ and $\frac{{ }^n{C}_{r+1}}{{ }^n{C}_{r+2}}=\frac{15}{70}$

$\Rightarrow \frac{ {\displaystyle \left(\frac{n!}{\left(n-r\right)!·r!}\right)}}{\left({\displaystyle \frac{n!}{\left(n-r-1\right)!·\left(r+1\right)!}}\right)}=\frac{2}{15}$ and $\frac{ {\displaystyle \left(\frac{n!}{\left(n-r-1\right)!·\left(r+1\right)!}\right)}}{\left({\displaystyle \frac{n!}{\left(n-r-2\right)!·\left(r+2\right)!}}\right)}=\frac{15}{70}$

$\Rightarrow \frac{\left(n-r-1\right)!·\left(r+1\right)!}{\left(n-r\right)!·r!}=\frac{2}{15}$ and $\frac{\left(n-r-2\right)!·\left(r+2\right)!}{\left(n-r-1\right)!·\left(r+1\right)!}=\frac{3}{14}$

$\Rightarrow \frac{\left(n-r-1\right)!·\left(r+1\right)·r!}{\left(n-r\right)·\left(n-r-1\right)!·r!}=\frac{2}{15}$ and $\frac{\left(n-r-2\right)!·\left(r+2\right)·\left(r+1\right)!}{\left(n-r-1\right)·\left(n-r-2\right)!·\left(r+1\right)!}=\frac{3}{14}$

$\Rightarrow \frac{ {\displaystyle r+1}}{n-r}=\frac{2}{15}$ and $\frac{ {\displaystyle r+2}}{n-r-1}=\frac{3}{14}$

$\Rightarrow 17r=2n-15$ and $17r=3n-31$

$\Rightarrow 3n-31=2n-15, \Rightarrow n=16$ and $r=1$

Hence, average $=\frac{{ }^n{C}_r{+}^n{C}_{r+1}{+}^n{C}_{r+2}}{3}$

$=\frac{{ }^{16}{C}_1{+}^{16}{C}_2{+}^{16}{C}_3}{3}$

$=232.$