Problem 2. In a sequence of six natural numbers, the third and each subsequent number is equal to the sum of the two preceding ones. Find the numbers if the fifth number is equal to 7.
Solution. Let the sequence of given numbers be $a, b, c, d, 7, e$. From the condition of the problem, we have:

$$
a+b=c, b+c=d, c+d=7, d+7=e
$$

from which

$$
a+b=d-b, a+2 b=d \text { and }(a+b)+(a+2 b)=7
$$

i.e.

$$
2 a+3 b=7
$$

This equation has a unique solution for $a=2, b=1$ and $c=3, d=4, e=11$.