1. The sequence of real numbers $a_{1}, a_{2}, a_{3}, \ldots, a_{100}$ satisfies the inequalities $a_{n}-2022 a_{n+1}+2021 a_{n+2} \geq 0 \quad$ for $\quad n=1,2,3, \ldots, 98$, and $\quad a_{99}-2022 a_{100}+2021 a_{1} \geq 0$, $a_{100}-2022 a_{1}+2021 a_{2} \geq 0$. Find $a_{22}$, if $a_{10}=10$. (10 points)
# Solution.

Let $b=a_{1}+a_{2}+a_{3}+\cdots+a_{100}, b_{n}=a_{n}-2022 a_{n+1}+2021 a_{n+2}, n=1,2,3, \ldots, 98$, $b_{99}=a_{99}-2022 a_{100}+2021 a_{1}, b_{100}=a_{100}-2022 a_{1}+2021 a_{2}$. By the condition $b_{n} \geq 0$, for $n=$ $1,2,3, \ldots, 100$. We have $b_{1}+b_{2}+b_{3}+\cdots+b_{100}=b-2022 b+2021 b=0 . \quad$ Therefore, $b_{n}=0$, for $n=1,2,3, \ldots, 100$. From this, we get $\left(a_{n}-a_{n+1}\right)+2021\left(a_{n+2}-a_{n+1}\right)=0, n=$ $1,2,3, \ldots, 98, a_{99}-a_{100}+2021\left(a_{1}-a_{100}\right)=0,\left(a_{100}-a_{1}\right)+2021\left(a_{2}-a_{1}\right)=0$.

We have $a_{2}-a_{3}=\frac{a_{1}-a_{2}}{2021}, \quad a_{3}-a_{4}=\frac{a_{1}-a_{2}}{2021^{2^{2}}}, \ldots, a_{99}-a_{100}=\frac{a_{1}-a_{2}}{2021^{1^{8}}}, a_{100}-a_{1}=\frac{a_{1}-a_{2}}{2021^{1^{9}}}$. Considering the equality $a_{100}-a_{1}=2021\left(a_{1}-a_{2}\right)$, we have $\frac{a_{1}-a_{2}}{2021^{99}}=2021\left(a_{1}-a_{2}\right)$. From this, we get $a_{n}=a_{1}$ for all $n=1,2,3, \ldots, 100$. Therefore, $a_{22}=a_{10}=10$.

## Answer: 10.