## Task B-4.2.

Marina, in a hurry, wrote only some of the numbers she was supposed to write in a $5 \times 5$ table. Confused, she explained to her friend that she would easily fill in the table later because the numbers in each row and each column form 5 consecutive terms of an arithmetic sequence. Fill in the table with the numbers Marina did not write.

| 3 |  |  |  |  |
| :--- | :--- | :--- | :--- | :--- |
|  |  |  |  |  |
|  |  | 31 |  |  |
|  | 11 |  |  |  |
|  |  |  |  | 27 |
## Solution.

Let the common difference of the arithmetic sequence in the first column be denoted by $d_{1}$, the common difference of the arithmetic sequence in the fifth row by $d_{2}$, and the fields in the table as $x$ and $y$ as shown in the figure:

| 3 |  |  |  |  |
| :---: | :---: | :---: | :---: | :---: |
|  |  |  |  |  |
| $3+2 d_{1}$ | $y$ | 31 |  |  |
|  | 11 |  |  |  |
| $x$ | $27-3 d_{2}$ |  |  | 27 |

The number $x$ is the fifth term of the arithmetic sequence in the first column and the first term of the arithmetic sequence in the fifth row. Therefore, $x=3+4 d_{1}$, or $x=27-4 d_{2}$. Then we have

$$
3+4 d_{1}=27-4 d_{2}
$$

from which it follows that $d_{1}+d_{2}=6$.

In the third row, the numbers $3+2 d_{1}, y, 31$ are three consecutive terms of an arithmetic sequence, so we have

$$
2 y=3+2 d_{1}+31
$$

or $y=17+d_{1}$.

The same applies to the numbers $y$, 11, and $27-3 d_{2}$ in the second column, so we have

$$
2 \cdot 11=y+27-3 d_{2}
$$

from which it follows that $y=-5+3 d_{2}$. Now we see that

$$
17+d_{1}=-5+3 d_{2}
$$

or $-d_{1}+3 d_{2}=22$, so we get the system of equations

$$
\begin{aligned}
d_{1}+d_{2} & =6 \\
-d_{1}+3 d_{2} & =22
\end{aligned}
$$

By adding the equations, we get $4 d_{2}=28$, so $d_{2}=7$, and then $d_{1}=6-7=-1$. We conclude that $x=-1, y=16$, and the table can now be easily filled in:

| 3 | 26 | 49 | 72 | 95 |
| :---: | :---: | :---: | :---: | :---: |
| 2 | 21 | 40 | 59 | 78 |
| 1 | 16 | 31 | 46 | 61 |
| 0 | 11 | 22 | 33 | 44 |
| -1 | 6 | 13 | 20 | 27 |