4.046. Three numbers form a geometric progression. If the second number is increased by 2, the progression becomes arithmetic, and if after that the last number is increased by 9, the progression becomes geometric again. Find these numbers.
## Solution.

Let $b_{1}, b_{2}, b_{3}$ be terms of a geometric progression, $b_{1}, b_{2}+2, b_{3}--$ be terms of an arithmetic progression, and $b_{1}, b_{2}+2, b_{3}+9$ be terms of a geometric progression. Then $b_{1}, b_{1} q+2, b_{1} q^{2}+9$ are terms of a geometric progression.

We have:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ 2 ( b _ { 1 } q + 2 ) = b _ { 1 } + b _ { 1 } q ^ { 2 } , } \\
{ ( b _ { 1 } q + 2 ) ^ { 2 } = b _ { 1 } \cdot ( b _ { 1 } q ^ { 2 } + 9 ) , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
2 b_{1} q+4=b_{1}+b_{1} q^{2}, \\
b_{1}^{2} q^{2}+4 b_{1} q+4=b_{1}^{2} q^{2}+9 b_{1},
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
b_{1}\left(q^{2}-2 q+1\right)=4, \\
b_{1}(9-4 q)=4 .
\end{array} \Rightarrow b_{1}=\frac{4}{9-4 q}, \frac{4\left(q^{2}-2 q+1\right)}{9-4 q}=4 \Rightarrow\right. \\
& \Rightarrow q^{2}+2 q-8=0, \text { from which } q_{1}=-4, q_{2}=2 .
\end{aligned}
$$

Then $\quad b_{1}^{\prime}=\frac{4}{9+16}=\frac{4}{25}, \quad b_{2}^{\prime}=\frac{4}{25}(-4)=-\frac{16}{25}, \quad b_{3}^{\prime}=\frac{64}{25} \quad$ or $b_{1}^{\prime \prime}=\frac{4}{9-8}=4, b_{2}^{\prime \prime}=4 \cdot 2=8, b_{3}^{\prime \prime}=16$.

Answer: 1) $4,8,16 ; 2) \frac{4}{25},-\frac{16}{25}, \frac{64}{25}$.