4.040. Three numbers form a geometric progression. If 4 is subtracted from the third number, the numbers will form an arithmetic progression. If 1 is then subtracted from the second and third terms of the resulting arithmetic progression, a geometric progression will again be obtained. Find these numbers.
Solution.

Let $b_{1}, b_{1} q, b_{1} q^{2}$ be the given numbers, then

$b_{1}, b_{1} q, b_{1} q^{2}$ - terms of a geometric progression,
$b_{1}, b_{1} q, b_{1} q^{2}-4$ - terms of an arithmetic progression,

$b_{1}, b_{1} q-1, b_{1} q^{2}-5$ - terms of a geometric progression.

Using the properties of the terms of an arithmetic progression (see No. 4.037) $2 a_{k}=a_{k-1}+a_{k+1}$ and the property of the terms of a geometric progression $b_{k}^{2}=b_{k-1} b_{k+1}, k=2,3, \ldots, n-1$, we get

$$
\left\{\begin{array}{l}
2 b_{1} q=b_{1}+b_{1} q^{2}-4 \\
\left(b_{1} q-1\right)^{2}=b_{1}\left(b_{1} q^{2}-5\right)
\end{array}\right.
$$

From the second equation, we have $b_{1}^{2} q^{2}-2 b_{1} q+1=b_{1}^{2} q^{2}-5 b_{1}, 2 b_{1} q-5 b_{1}=1$, $b_{1}=\frac{1}{2 q-5}$. Substituting this value of $b_{1}$ into the first equation, we get $\frac{2 q}{2 q-5}=\frac{1+q^{2}}{2 q-5}-4$, from which $q^{2}-10 q+21=0$, from which $q_{1}=7, q_{2}=3$.

Then $b_{1}^{\prime}=\frac{1}{9}, b_{2}^{\prime}=\frac{7}{9}, b_{3}^{\prime}=\frac{49}{9}$ or $b_{1}^{\prime \prime}=1, b_{2}^{\prime \prime}=3, b_{3}^{\prime \prime}=9$.

Answer: 1$\left.) \frac{1}{9}, \frac{7}{9}, \frac{49}{9} ; 2\right) 1,3,9$.