We have an 8-term arithmetic sequence and a 4-term geometric sequence. The first term in both sequences is 1; the last terms are also equal to each other. What are these sequences if the sum of the four terms of the geometric sequence is 21 more than the eighth term of the arithmetic sequence?
In the sense of the task:

$$
1+7 d=q^{3}
$$

and

$$
1+q+q^{2}+q^{3}=1+7 d+21
$$

or, taking into account the first equation:

$$
1+q+q^{2}=21
$$

from which

$$
q=4 \quad \text { and } \quad q'=-5
$$

substituting into the first equation,

$$
d=9 \quad \text { and } \quad d'=-18
$$

The sought progressions are therefore:

$1,10,19, \ldots 64 ; 1,4,16,64$

$1,-17,-35, \ldots-125 ; 1,-5,25,-125$.

(Ferencz Stromfeld, Budapest.)

The problem was also solved by: Dolowschiák M., Filkorn J., Freibauer E., Jancsó G., Kárf J., Kohn B., Krausz B., Krisztián Gy., Kürth A., Lukhaub Gy., Obláth R., Ovenden S., Petrogalli G., Raab M., Rehberger Z., Schneider E., Spitzer Ö., Weisz Á., Weisz J., Wieszner B.