There is an 8-term arithmetic progression and a 4-term geometric progression. The first term of each progression is 1, and their last terms are equal. The sum of the geometric progression is 21 greater than the last term of the arithmetic progression. The two progressions need to be determined.
We denote the common ratio of the geometric progression by $q$, and the common difference of the arithmetic progression by $d$; according to the problem:

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

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

From (2), $q_{1}=4, q_{2}=-5$, substituting these values into (1): $d_{1}=9, d_{2}=-18$.

The problem was solved by: Feuer Mór, Freund Antal, Friedmann Bernát, Geist Emil, Goldstein Zsigmond, Grünhut Béla, Hönigsfeld Tivadar, Hofbauer Ervin, Kántor Nándor, Klein Mór, Kornis Ödön, Piovarcsy Jenő, Riesz Frigyes, Roth Miksa, Schneider Béla, Tauszig Aladár, Weisz Ármin.