Three numbers form a geometric sequence. Their sum is 19; by reducing the last number by 1, we get an arithmetic sequence; which are these three numbers?
Let $a$, $a q$ and $a q^{2}$ be the three numbers we are looking for.

According to the problem,

$$
a\left(q^{2}+q+1\right)=19
$$

and

$$
\frac{3}{2}\left(a+a q^{2}-1\right)=18
$$

From these two equations, we derive that

$$
a_{1}=4, a_{2}=9, q_{1}=\frac{3}{2} \text { and } q_{2}=\frac{2}{3}
$$

thus, the numbers we are looking for are $4,6,9$ and $9,6,4$.

(Popoviciu Miklós, Brassó.)

Number of solutions: 48.