1085*. Several natural numbers form an arithmetic progression, starting with an even number. The sum of the odd terms of the progression is 33, and the sum of the even terms is 44. Find the progression and the number of its terms. List all solutions.
Let's denote the first term of the arithmetic progression by $2a$ (where $a \in \mathbb{N}$), the common difference by $d$, and the number of terms by $n$. We apply the formula for the sum of the first $n$ terms of an arithmetic progression:

$$
S_{n}=\frac{2a_{1}+d(n-1)}{2} n
$$

For the sum of the first $n$ terms of the arithmetic progression:

$$
\frac{4a+d(n-1)}{2} n=77, \quad (4a+d(n-1)) n=2 \cdot 7 \cdot 11
$$

We consider two cases.

1) Suppose $n$ is odd.

Then $n$ is equal to 7, 11, or 77. The last possibility is immediately ruled out because in this case the sum in parentheses would be 2, which is impossible.

Let $n=7$. We get:

$$
(4a+6d) \cdot 7=2 \cdot 7 \cdot 11, \quad 2a+3d=11
$$

From this, either $a=1, d=3$, or $a=4, d=1$. We have two arithmetic progressions:

$$
\begin{aligned}
& 2 ; 5 ; 8 ; 11 ; 14 ; 17 ; 20 \\
& 8 ; 9 ; 10 ; 11 ; 12 ; 13 ; 14
\end{aligned}
$$

For each of them, the sum of the odd terms is 33, and thus the sum of the even terms is 44, so both satisfy the conditions of the problem.

Let $n=11$. Then

$$
(4a+6d) \cdot 11=2 \cdot 7 \cdot 11, \quad 2a+3d=7
$$

From this, $a=2, d=1$. We get a progression with 11 terms:

$$
4 ; 5 ; 6 ; \ldots ; 14
$$

But here the sum of the odd terms is greater than 33:

$$
5+7+9+11+13=45>33
$$

Therefore, this case is impossible.

2) Suppose $n$ is even.

Then $n$ is equal to 14, 22, or 154. Check independently that in none of these three cases there are solutions.

Answer: $2 ; 5 ; 8 ; 11 ; 14 ; 17 ; 20$ or $8 ; 9 ; 10 ; 11 ; 12 ; 13 ; 14$. The number of terms in both progressions is 7.