If the coefficients of $x^9, x^{10}, x^{11}$ in the expansion of ${\left(1+x\right)}^n$ are in A.P then prove that $n^2-41n+398=0$ 
 

Given expression is ${\left(1+x\right)}^n$

the coefficients of $x^9, x^{10}, x^{11}$are $n_{C_9}, n_{C_{10}}, n_{C_{11}}$respectively.

 Given $n_{C_9}, n_{C_{10}}, n_{C_{11}}$are in A.P 
$$\begin{gathered} 2{(}^n{C}_{10}){=}^n{C}_9{+}^n{C}_{11} \\ \Rightarrow 2=\frac{{ }^n{C}_9}{{ }^n{C}_{10}}+\frac{{ }^n{C}_{11}}{{ }^n{C}_{10}} \end{gathered}$$

$We know that \frac{{ }^n{C}_r}{{ }^n{C}_{r-1}}=\frac{n-r+1}{r}$

$2=\frac{10}{n-10+1}+\frac{n-11+1}{11}$

$$\begin{gathered} 2=\frac{10}{n-9}+\frac{n-10}{11} \\ \Rightarrow 22(n-9)=110+(n-9)(n-10) \\ \Rightarrow 22n-198=110+n^2-19n+90 \\ \Rightarrow n^2-41n+398=0 \end{gathered}$$

$n^2-41n+398=0$