4.064. Three numbers, of which the third is 12, form a geometric progression. If 12 is replaced by 9, the three numbers form an arithmetic progression. Find these numbers.
## Solution.

Let $b_{1}, b_{2}, 12$ be terms of a geometric progression, and $\mathrm{a} b_{1}, b_{2}, 9$ be terms of an arithmetic progression. Then

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ b _ { 1 } q ^ { 2 } = 1 2 , } \\
{ b _ { 1 } + 2 d = 9 , } \\
{ b _ { 1 } q = b _ { 1 } + d , }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ ( 9 - 2 d ) q ^ { 2 } = 1 2 , } \\
{ b _ { 1 } = 9 - 2 d , } \\
{ ( 9 - 2 d ) ( q - 1 ) = d , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(9-2 d) \cdot \frac{(9-d)^{2}}{(9-2 d)^{2}}=12 \\
b_{1}=9-2 d \\
q=\frac{9-d}{9-2 d}
\end{array} \Leftrightarrow\right.\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \frac { ( 9 - d ) ^ { 2 } } { 9 - 2 d } = 1 2 , } \\
{ b _ { 1 } = 9 - 2 d , } \\
{ q = \frac { 9 - d } { 9 - 2 d } , }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ d _ { 1 } = - 9 } \\
{ d _ { 2 } = 3 , } \\
{ b _ { 1 } = 9 - 2 d , } \\
{ q = \frac { 9 - d } { 9 - 2 d } , }
\end{array} \Leftrightarrow \left[\begin{array}{l}
\left\{\begin{array}{l}
d_{1}=-9, \\
b_{1}^{\prime}=9-2(-9)=27 \\
q_{1}=\frac{9+9}{9+18}=\frac{2}{3}
\end{array}\right. \\
\left\{\begin{array}{l}
d_{2}=3 \\
b_{1}^{\prime \prime}=9-6=3
\end{array}\right. \\
q_{2}=\frac{9-3}{9-6}=2 .
\end{array}\right.\right.\right.
\end{aligned}
$$

In the first case, $b_{1}^{\prime}=27, b_{2}^{\prime}=b_{1} q=27 \cdot \frac{2}{3}=18, b_{3}^{\prime}=12$; in the second case, $b_{1}^{\prime \prime}=3, b_{2}^{\prime \prime}=b_{1} q=3 \cdot 2=6, b_{3}^{\prime \prime}=12$.

Answer: 1) 27, 18, 12; 2) $3,6,12$.