10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=p, a_{2}=p+1, \\
a_{n+2}-2 a_{n+1}+a_{n}=n-20,
\end{array}
$$

where $p$ is a given real number, and $n$ is a positive integer. Try to find the value of $n$ that minimizes $a_{n}$.
10. Let $b_{n}=a_{n+1}-a_{n}$.

From the problem, we have $b_{n+1}-b_{n}=n-20$, and $b_{1}=1$. Then, $b_{n}-b_{1}=\sum_{i=1}^{n-1}\left(b_{i+1}-b_{i}\right)=\sum_{i=1}^{n-1}(i-20)$.
Thus, $b_{n}=\frac{(n-1)(n-40)}{2}+1$.
Also, $a_{3}=a_{2}+b_{2}=p-17<a_{1}<a_{2}$, so when the value of $a_{n}$ is the smallest, we have $n \geqslant 3$, and
$$
\begin{array}{l}
\left\{\begin{array}{l}
a_{n} \leqslant a_{n+1}, \\
a_{n} \leqslant a_{n-1},
\end{array}\right. \\
\left\{\begin{array}{l}
b_{n}=\frac{(n-1)(n-40)}{2}+1 \geqslant 0, \\
b_{n-1}=\frac{(n-2)(n-41)}{2}+1 \leqslant 0 .
\end{array}\right.
\end{array}
$$

Therefore, when $n=40$, the value of $a_{n}$ is the smallest.