In a quarry, 50 stone blocks have been carved out. The stones can be arranged in such a way that in the row - starting from the second one - each stone weighs $2 \mathrm{~kg}$ more than the one in front of it. The weight of the first stone is $370 \mathrm{~kg}$. Can all the stone blocks be transported with $7 \mathrm{db} 3$ ton trucks, in a single trip? (Overloading the trucks is not allowed.)
The stones cannot be transported under the given conditions (of course, each one must remain intact). Indeed, out of the 7 trucks, at least one would have to carry 8 stones, and even the lightest 8 stones would overload this truck. Their total weight can simply be obtained by adding 8 times the weight of the first stone to the sum of the excess weights of the next 7 stones:

$$
8 \cdot 370 + (2 + 4 + 6 + 8 + 10 + 12 + 14) = 3016 \text{ kg} > 3 \text{ t}.
$$

Remarks. 1. The sum $2 + 4 + 6 + 8 + 10 + 12 + 14$ at the end of the solution can be calculated more quickly by noting that pairing the terms from the front and back, the sums $2 + 14, 4 + 12, 8 + 10$ are equal and their common value is twice the middle term 8. Thus, the value of the sum is $7 \cdot 8 = 56$.

A similar observation can be used to quickly determine the total weight of the 50 stone bales. However, this is not necessary for solving the problem, and this result might even be misleading in a certain sense, as it shows that the total weight of the stone bales does not exceed the combined capacity of the trucks ($20950 \text{ kg}$).

2. It can be easily shown that if I allowed even a single stone bale to be cut in half, it would already be possible to transport them in a single trip.