The first three terms of a geometric sequence form a geometric sequence with a common ratio of 2, while the common ratios of the three geometric sequences form an arithmetic sequence with a common difference of 1. The sum of the second terms of the three geometric sequences is 24, and the sum of the first three terms of the third geometric sequence is 84. The three geometric sequences need to be established.
The three geometric sequences can be written in the following form:

$$
\begin{gathered}
a, a(q-2), a(q-2)^{2}, \ldots \\
2 a, 2 a(q-1), 2 a(q-1)^{2}, \ldots \\
4 a, 4 a q, 4 a q^{2}, \ldots
\end{gathered}
$$

According to the problem:

$$
7 a q-4 a=24
$$

and

$$
4 a\left(1+q+q^{2}\right)=84
$$

Substituting the value of $a$ from (1) into (2):

$$
8 q^{2}-41+36=0
$$

from which

$$
q=4, q^{\prime}=\frac{9}{8}
$$

and

$$
a_{1}=1, a_{1}^{\prime}=\frac{192}{31}
$$

Thus, the sought progressions are

$$
\begin{gathered}
1,2,4 \ldots ; 2,6,18 \ldots 4,16,64 \ldots \\
\frac{192}{31}, \frac{168}{31}, \ldots \frac{384}{31}, \frac{48}{31}, \ldots ; \frac{768}{31}, \frac{864}{31}, \frac{972}{31}, \ldots
\end{gathered}
$$

(Aurél Erdös.)

Number of solutions: 66.