4.017. Find four numbers that form a geometric progression, where the third term is 9 more than the first, and the second term is 18 more than the fourth.
## Solution.

From the condition, we have $\left\{\begin{array}{l}b_{3}-b_{1}=9, \\ b_{2}-b_{4}=18 .\end{array}\right.$

Using formula (4.6), we get

$$
\left\{\begin{array} { l } 
{ b _ { 1 } q ^ { 2 } - b _ { 1 } = 9 , } \\
{ b _ { 1 } q - b _ { 1 } q ^ { 3 } = 1 8 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}\left(q^{2}-1\right)=9, \\
-b_{1} q\left(q^{2}-1\right)=18
\end{array} \Rightarrow \frac{q^{2}-1}{-q\left(q^{2}-1\right)}=\frac{9}{18} \Rightarrow\right.\right.
$$

$\Rightarrow q=-2$. Substituting $q=-2$, we get $b_{1}=\frac{9}{q^{2}-1}=\frac{9}{4-1}=3$. Then $b_{2}=-6, b_{3}=12, b_{4}=-24$.

Answer: $3,-6,12,-24$.