Let's find a 10-term arithmetic sequence consisting entirely of prime numbers.
Let's denote the difference between consecutive terms of the sought arithmetic sequence by $d$. If $d$ were odd, then every second term of the sequence would be even, and the terms could not be prime numbers. (We are, of course, considering only integers as terms.) Therefore, $d$ must be divisible by 2. If $d$ were not divisible by 3, then among any three consecutive terms of the sequence, one would always be divisible by 3, so $d$ must also be divisible by 3. Similarly, it can be shown that $d$ is divisible by 5 and 7, so $d$ is divisible by $2 \cdot 3 \cdot 5 \cdot 7 = 210$.

In order to work with as small (positive) numbers as possible, we look for a sequence with a difference of $d = 210$. We will examine the primes that can be the starting term $a$ of the sequence in increasing order until we find one for which all terms $(a + 210i)$ for $(0 \leq i \leq 9)$ are prime. It is clear that $a \geq 11$.

We expect to find a solution where every term is less than 2500, so it is sufficient to ensure that the subsequent terms $(1 \leq i \leq 9)$ are not divisible by any prime less than $\sqrt{2500} = 50$. The starting term itself can only be divisible by such a prime if it is equal to it.

We can work with smaller numbers if, for successive values of $p$, we consider the remainder $m$ of the division $(a : p)$ instead of $a$ and the remainder $r$ of the division $(210 : p)$ instead of $d$, i.e., $a = \alpha p + m, 210 = \delta p + r$, where $\alpha, \delta$ are natural numbers, $0 \leq m < p$ and $0 < r < p$. Thus, we consider the following 10 numbers:

$$
m, m + r, m + 2r, \ldots, m + 9r
$$

since $m + ri = (a + 210i) - (\alpha + \delta i) \cdot p$ is divisible by $p$ if and only if $(a + 210i)$ is divisible by it.

For each of the above $p$ values, we collect in a table those values of $m$ for which (1) none of the terms is divisible by $p$. The table below contains these values for $11 \leq p \leq 19$. (For $p \geq 23$, it is shorter to list the values of $m$ that are not allowed.) - For example, for $p = 11$, $r = 1$, and $m = 1$ is still valid because with it (1) consists of the numbers $1, 2, \ldots, 10$, but for $m \geq 2$, a number divisible by 11 would appear among them.

| $p$ | $r$ | Possible values of $m$: |  |  |  |  |  |  |  |  |  |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 11 | 1 | 0, | 1; |  |  |  |  |  |  |  |  |
| 13 | 2 | 0, | 2, | 4, | 6; |  |  |  |  |  |  |
| 17 | 6 | 0, | 6, | 12, | 1, | 7, | 13, | 2, | 8; |  |  |
| 19 | 1 | 0, | 1, | 2, | 3, | 4, | 5, | 6, | 7, | 8, | 9; |


| $p$ | $r$ | Prohibited values of $m$: |  |  |  |  |  |  |  |  |  |
| :--- | ---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 23 | 3 | 20, | 17, | 14, | 11, | 8, | 5, | 2, | 22, | 19; |  |
| 29 | 7 | 22, | 15, | 8, | 1, | 23, | 16, | 9, | 2, | 24; |  |
| 31 | 24 | 7, | 14, | 21, | 28, | 4, | 11, | 18, | 25, | 1; |  |
| 37 | 25 | 12, | 24, | 36, | 11, | 23, | 35, | 10, | 22, | 34; |  |
| 41 | 5 | 36, | 31, | 26, | 21, | 16, | 11, | 6, | 1, | 37; |  |
| 43 | 38 | 5, | 10, | 15, | 20, | 25, | 30, | 35, | 40, | 42; |  |
| 47 | 22 | 25, | 3, | 28, | 6, | 31, | 9, | 34, | 12, | 37. |  |

Based on these, the starting number of the sequence cannot be 11, because it gives a prohibited remainder for $p = 13$, which is not present in the 13th row (and is present in the 23rd row). The further search is significantly simplified by taking as a basis that $a$ gives a remainder of 1 when divided by 11. The first such prime number is 23, but it also gives a prohibited remainder for $p = 13$. Similarly, $a = 67, 89$ give prohibited remainders for $p = 17, 13$, and 45, 111, 133, 155, and 177 are not prime. The number $a = 199$ gives remainders of 1, 4, 12, 9, 15, 25, 13, 14, 35, 27, 11 for the above primes, all of which are allowed, and $a = 199$ itself is prime, so the first ten terms of the arithmetic sequence with starting term $a = 199$ and difference $d = 210$ are certainly prime.

Remarks. 1. The number $(-11)$ can also be inserted at the beginning of the sequence, but in such problems, it is customary to focus on positive numbers.

2. With less thought and preparation, it would be simpler to take the terms of the sequence $1 + 22k$ as $a$, but then it would be necessary to individually check whether every term is prime in the sequences.