We know the following about an arithmetic and a geometric sequence: a) their terms are different natural numbers; b) their first term is 1; c) the sum of the first 4 terms is equal in both sequences. Write down four such pairs of sequences using possibly small numbers.
Let the difference of the arithmetic sequence be $d$, and the common ratio of the geometric sequence be $q$, both being natural numbers greater than 1. Thus, according to the condition c),

$$
\begin{gathered}
4+6 d=1+q+q^{2}+q^{3} \\
d=\frac{q\left(q^{2}+q+1\right)-3}{6}
\end{gathered}
$$

This is always positive for the considered values of $q$, and it is an integer if the numerator is divisible by both 2 and 3. The first condition is satisfied if $q$ is an odd number, because then the expression in the parentheses is also odd. The second condition is satisfied if $q$ is divisible by 3, or if $q^{2}+q+1=(q-1)^{2}+3 q$, meaning if $(q-1)^{2}$ is divisible by 3. The first condition holds for numbers of the form $q=3M$, and the second for numbers of the form $q=3M+1$, where $M$ is an integer, since a square number is divisible by 3 if and only if the base number is divisible by 3, due to the primality of 3.

Combining the two conditions, the remainder of the numerator when divided by 6 must be odd, and it must be one of the numbers 0 and 3, or 1 and 4. Only the remainders 1 and 3 satisfy both conditions, so the solution to the problem is every natural number greater than 1 of the form $q=6k+3$ and $6k+1$, and the corresponding $d$ can be calculated from (1).

The smallest four solutions are:

For $q=3, d=6$: $1+7+13+19=1+3+9+27=40$

For $q=7, d=66$: $1+67+133+199=1+7+49+343=400$

For $q=9, d=136$: $1+137+273+409=1+9+81+729=820$

For $q=13, d=396$: $1+397+793+1189=1+13+169+2197=2380$.

Agnes Juhasz (Budapest, Berzsenyi D. Gymnasium, 2nd grade)