Given that coefficient of $x^9, x^{10}$ and $x^{11}$ in the expansion of $(1+n)^n$ are in A.P.
It means ${ }^n C_9 \cdot{ }^n C_{10} \cdot{ }^n C_{11}$ are in A.P.
$\begin{aligned}
& \therefore \quad 2^n C_{10}={ }^n C_9+{ }^n C_{11} \\
& 2=\frac{{ }^n C_9}{{ }^n C_{10}}+\frac{{ }^n C_{11}}{{ }^n C_{10}} \\
& \Rightarrow \quad 2=\frac{n !}{9 !(n-9) !} \times \frac{10 !(n-10) !}{n !}
\end{aligned}$
$$
+\frac{n !}{11 !(n-11) !} \times \frac{10 !(n-10) !}{n !}
$$
$\begin{aligned} & \Rightarrow \quad 2=\frac{10}{n-9}+\frac{n-10}{11} \\ & 2=\frac{110+(n-10)(n-9)}{11(n-9)} \\ & 2=\frac{110+n^2-9 n-10 n+90}{11 n-99} \\ & 22 n-198=200+n^2-19 n \\ & \therefore \quad n^2-41 n=-398\end{aligned}$