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 $x_{n}=\frac{x_{n-1}+298 x_{n}+x_{n+1}}{300}$ holds. Find $\sqrt{\frac{x_{2023}-x_{2}}{2021} \cdot \frac{2022}{x_{2023}-x_{1}}}-2023$.
# 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,

$$
\begin{gathered}
x_{n}=\frac{x_{n-1}+298 x_{n}+x_{n+1}}{300} \\
2 x_{n}=x_{n-1}+x_{n+1} \\
x_{n}-x_{n-1}=x_{n+1}-x_{n}
\end{gathered}
$$

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

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

Answer. -2022.