Out of five numbers, the first four form an arithmetic progression, the sum of which is 40. The last three form a geometric progression, in which the product of the two outer terms is 32 times the second of the five numbers. The five numbers are to be determined.
The five numbers can be written as:

$$
a-d, a, a+d, a+2 d, \frac{(a+2 d)^{2}}{a+d}
$$

According to the problem, therefore:

$$
4 a+2 d=40
$$

and

$$
\frac{(a+2 d)^{2}}{a+d}(a+d)=32 a
$$

or

$$
(a+2 d)^{2}=32 a
$$

Substituting $2 d=40-4 a$ from (1):

$$
9 a^{2}-272 a+1600=0
$$

from which

$$
a=\frac{200}{9}, \quad a^{\prime}=8
$$

thus

$$
d=-\frac{220}{9}, \quad d^{\prime}=4
$$

and so the five numbers are:

$$
\frac{420}{9}, \frac{200}{9},-\frac{20}{9},-\frac{240}{9},-320
$$

and

$4,8,12,16, \frac{64}{3}$.

(Aurél Erdös.)

Number of solutions: 53.