Three numbers form a geometric progression. If we add 8 to the second number, an arithmetic progression is formed. If we add 64 to the third term of this arithmetic progression, we again get a geometric progression. What are the original numbers?
Let the three numbers be: $a, a q, a q^{2}$.

According to the problem:

$$
2(a q+8)=a+a q^{2}
$$

and

$$
(a q+8)^{2}=a\left(a q^{2}+64\right)
$$

(1) can also be written as

$$
a\left(1-2 q+q^{2}\right)=16
$$

Simplifying and rearranging (2):

$$
a(4-q)=4
$$

The ratio of (3) and (4) is

$$
\frac{1-2 q+q^{2}}{4-q}=4
$$

from which

$$
q^{2}+2 q-15=0
$$

hence

$$
q_{1}=3, q_{2}=-5
$$

and thus

$$
a_{1}=4 \quad \text { and } \quad a_{2}=\frac{4}{9}
$$

Therefore, the three numbers are 4, 12, 36 or $\frac{4}{9},-\frac{20}{9}, \frac{100}{9}$.

Balatoni Ferenc (Bp. III., Árpád g. III. o. t.)