2. Variant 1.

Find the number of four-digit numbers for which the last three digits form an increasing arithmetic progression (numbers cannot start with zero).
Answer: 180.

Solution: The difference of the progression cannot be greater than 4.

For $d=1$ - there are eight suitable progressions of digits: from 012 to 789.

For $d=2$ - six: from 024 to 579.

For $d=3$ - four: from 036 to 369.

For $d=4$ - two: 048 and 159.

In total, $8+6+4+2=20$ options for the last three digits. By choosing the first digit in nine ways for each of them, we get the answer $20 \cdot 9=180$.