The sum of a certain infinite geometric progression is greater than the sum of the infinite geometric progression $\frac{5}{12}, \frac{5}{16} \ldots$ by its own constant ratio; the first term of the former progression is equal to the first term of the given progression. Calculate the ratio and sum of both progressions.
The given progression has a common ratio of $\frac{3}{4}$, and its sum is $\frac{5}{3}$; therefore, according to the problem,

$$
\frac{5}{3}+q=\frac{5}{12} \cdot \frac{1}{1-q}
$$

or

$$
q^{2}+\frac{2}{3} q-\frac{5}{4}=0
$$

from which

$$
q=\frac{5}{6}, q^{\prime}=-\frac{3}{2}
$$

and thus

$$
S=2.5, S^{\prime}=\frac{1}{6}
$$

(Béla Bayer, Losoncz.)

The problem was solved by: Baumann J., Beck I., Czank K., Demeter J., Faith F., Fekete N., Filkorn J., Kéler E., Kerekes T., Kőnig D., Krausz B., Krisztián Gy., Lukhaub Gy., Lupsa Gy., Messik G., Messik V., Perl Gy., Póka Gy., Russo M., Sasvári G., Sasvári J., Scheuer R., Schwarz J., Smodics K., Weisz A.