1. Let the arithmetic progression be denoted by $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.
2. Given the condition $a_n + a_{n+1} = a_1 + a_2 + \cdots + a_{3n-1}$, we need to express both sides in terms of $a$ and $d$.

3. The left-hand side (LHS) is:
$$a_n + a_{n+1} = \left(a + (n-1)d\right) + \left(a + nd\right) = 2a + (2n-1)d$$

4. The right-hand side (RHS) is the sum of the first $3n-1$ terms of the arithmetic progression:
$$a_1 + a_2 + \cdots + a_{3n-1} = \sum_{k=1}^{3n-1} \left(a + (k-1)d\right)$$
This sum can be calculated using the formula for the sum of an arithmetic series:
$$S_{3n-1} = \frac{(3n-1)}{2} \left(2a + (3n-2)d\right)$$

5. Equating the LHS and RHS, we get:
$$2a + (2n-1)d = \frac{(3n-1)}{2} \left(2a + (3n-2)d\right)$$

6. Simplifying the equation:
$$4a + 2(2n-1)d = (3n-1)(2a + (3n-2)d)$$
$$4a + 4nd - 2d = 6na + (3n-1)(3n-2)d$$
$$4a + 4nd - 2d = 6na + (9n^2 - 6n - 3n + 2)d$$
$$4a + 4nd - 2d = 6na + (9n^2 - 9n + 2)d$$

7. Rearranging terms to isolate $a$ and $d$:
$$4a - 6na = (9n^2 - 9n + 2)d - 4nd + 2d$$
$$4a(1 - \frac{3n}{2}) = d(9n^2 - 13n + 2)$$
$$4a(1 - \frac{3n}{2}) = d(9n^2 - 13n + 2)$$

8. Solving for $d$:
$$d = \frac{4a(1 - \frac{3n}{2})}{9n^2 - 13n + 2}$$

9. We need to prove that there are no zero terms in the progression. For any term $a_m = a + (m-1)d$ to be zero:
$$a + (m-1)d = 0$$
Substituting $d$:
$$a + (m-1) \frac{4a(1 - \frac{3n}{2})}{9n^2 - 13n + 2} = 0$$
$$a \left(1 + (m-1) \frac{4(1 - \frac{3n}{2})}{9n^2 - 13n + 2}\right) = 0$$

10. Since $a \neq 0$, the term inside the parentheses must be non-zero:
$$1 + (m-1) \frac{4(1 - \frac{3n}{2})}{9n^2 - 13n + 2} \neq 0$$

11. Simplifying the expression inside the parentheses:
$$1 + (m-1) \frac{4 - 6n}{9n^2 - 13n + 2} \neq 0$$

12. Since $9n^2 - 13n + 2 \neq 0$ for all natural $n$, and $4 - 6n \neq 0$ for all natural $n$, the term inside the parentheses is non-zero.

13. Therefore, there are no zero terms in the arithmetic progression.

$\blacksquare$