6.24 Find four numbers, the first three of which form a geometric progression, and the last three form an arithmetic progression. The sum of the extreme numbers is 21, and the sum of the middle numbers is 18.
6.24 Let $a_{1}, a_{2}, a_{3}, a_{4}$ be the required numbers. Using the conditions, as well as formulas (6.4) and (6.8), we obtain the system

$$
\left\{\begin{array}{l}
a_{2}^{2}=a_{1} a_{3} \\
2 a_{3}=a_{2}+a_{4} \\
a_{1}+a_{4}=21 \\
a_{2}+a_{3}=18
\end{array}\right.
$$

By eliminating $a_{1}, a_{2}, a_{4}$ from these equations, we have $4 a_{3}^{2}-75 a_{3}+324=0$, from which we find $a_{3}=12$ or $a_{3}=6.75$.

Answer: $3 ; 6 ; 12 ; 18$ or $18.75 ; 11.25 ; 6.75 ; 2.25$.