How Many Ways to Fill a 3x3 Grid with Arithmetic Progressions?

We know that in any arithmetic progression, the middle term is the average of the first and the last term. So, let us fix the middle term for each row and column, and then find the possible arrangements of the other two numbers.

There are 9 choices for the middle number. Let's say we choose 5. Then the first and last numbers in the row/column must add up to 10, in order for them to have an average of 5. We can only choose pairs that add up to 10 in the following ways: 1 and 9, 2 and 8, 3 and 7, 4 and 6, and 6 and 4, 7 and 3, 8 and 2, 9 and 1. So, there are 8 choices for the first number and 1 choice for the last number (since it is determined by the first number). This gives us a total of 8 ways to arrange the numbers in that row/column.

Therefore, there are 9 choices for the middle number, and for each choice, there are 8 ways to arrange the other two numbers. This gives us a total of $9\times8=\boxed{72}$ ways to fill the grid.