Task 4. (20 points) For the numerical sequence $\left\{x_{n}\right\}$, all terms of which, starting from $n \geq 2$, are distinct, the relation $x_{n}=\frac{x_{n-1}+198 x_{n}+x_{n+1}}{200}$ holds. Find $\sqrt{\frac{x_{2023}-x_{1}}{2022} \cdot \frac{2021}{x_{2023}-x_{2}}}+2022$.
# Solution.

From the given relations in the problem, it is easily deduced that for all $n \geq 2$, $x_{n}-x_{n-1}=x_{n+1}-x_{n}$, which implies that the sequence is an arithmetic progression. Indeed,

$$
\begin{gathered}
x_{n}=\frac{x_{n-1}+198 x_{n}+x_{n+1}}{200} \\
2 x_{n}=x_{n-1}+x_{n+1} \\
x_{n}-x_{n-1}=x_{n+1}-x_{n}
\end{gathered}
$$

Let the common difference of this progression be $d, d \neq 0$ (as per the condition).

Then $\sqrt{\frac{x_{2023}-x_{1}}{2022} \cdot \frac{2021}{x_{2023}-x_{2}}}+2022=\sqrt{\frac{x_{1}+2022 d-x_{1}}{2022} \cdot \frac{2021}{x_{1}+2022 d-x_{1}-d}}+2022=$ $=\sqrt{\frac{2022 d}{2022} \cdot \frac{2021}{2021 d}}+2022=1+2022=2023$.

Answer. 2023.