## Task 34/70

In an arithmetic sequence of the 1st order, it is known that all terms are non-negative integers. Exactly 1 term is a single-digit number, 9 terms are two-digit numbers, 81 are three-digit numbers, and 819 are four-digit numbers. Determine the sequence!
The sequence is $\left\{a_{i}\right\}$ with $i=1 ; 2 ; \ldots ; 910$. Since all $a_{i}$ are integers, $d$ is also an integer, and if $a_{1}$ is the smallest number in the sequence (which can be assumed without loss of generality), then $d>0$. According to the formation rule of the arithmetic sequence of the first order, $a_{i}=a_{1}+(i-1) d$. From the information given in the problem, the following inequalities can be established:

$$
\begin{array}{r}
a_{2}=a_{1}+d \geq 10 \\
a_{10}=a_{1}+9 d \leq 99 \\
a_{92}=a_{1}+91 d \geq 1000
\end{array}
$$

From (1) and (2), it follows that $8 d \leq 89, d \leq 11$. From (2) and (3), it follows that $82 d \geq 901, d>10$. Therefore, $d=11$. From (2), with this value $d=11$, it immediately follows that $a_{1} \leq 0$; however, since all terms are non-negative, $a_{1}=0$. Thus, the sequence $\left\{a_{i}\right\}=\{11(i-1)\}$ with $i=1 ; 2 ; \ldots, 910$ is the desired one.