1. Determine all possible fillings of the empty fields in the table so that the numbers in each row and each column represent an arithmetic progression.
Solution. Let's denote by $a_{ij}$ the element that is in the $i$-th row and $j$-th column, with $r_{1}$ being the difference of the arithmetic progression in the first row and $k_{1}$ the difference of the arithmetic progression in the first column.

(1) From the first row, we have $14 - a_{11} = 4r_{1}$, and from the first column, $2 - a_{11} = 4k_{1}$. By subtracting these equations, we get $4r_{1} - 4k_{1} = 12$, i.e., $r_{1} - k_{1} = 3$.

(2) Since there are two numbers near $a_{23}$, we will express this element in two different ways.

From the third column, we have $a_{23} = \frac{a_{13} + a_{33}}{2} = \frac{14 - 2r_{1} + 20}{2} = 17 - r_{1}$.

From the second row, we have $\frac{a_{23} + a_{21}}{2} = a_{22}$, i.e., $\frac{a_{23} + 2 - 3k_{1}}{2} = 10$, from which $a_{23} = 18 + 3k_{1}$.

By equating the two expressions for $a_{23}$, we get $r_{1} + 3k_{1} = -1$.

(3) From (1) and (2), we have the system of equations for the differences of the arithmetic progressions in the first row and the first column:

$$
\left\{\begin{array}{l}
r_{1} - k_{1} = 3 \\
r_{1} + 3k_{1} = -1
\end{array} \Leftrightarrow \left\{\begin{array}{l}
r_{1} - k_{1} = 3 \\
4k_{1} = -4
\end{array} \Leftrightarrow \left\{\begin{array}{l}
r_{1} = 2 \\
k_{1} = -1
\end{array}\right.\right.\right.
$$

Now we can fill in the first row and the first column, and then easily fill in the remaining

| 6 | 8 | 10 | 12 | 14 |
| :--- | :--- | :--- | :--- | :--- |
| 5 | 10 | 15 | 20 | 25 |
| 4 | 12 | 20 | 28 | 36 |
| 3 | 14 | 25 | 36 | 47 |
| 2 | 16 | 30 | 44 | 58 |

cells in the table. The final solution: