4.044. 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.
## Solution.

Let $b_{1}, b_{1} q, b_{1} q^{2}, b_{1} q^{2}+d$ be the given numbers.

From the condition, we have:

$b_{1}, b_{1} q, b_{1} q^{2}$ are terms of a geometric progression,

$b_{1}, b_{1} q+d, b_{1} q+2 d$ are terms of an arithmetic progression,

$b_{1}+b_{1} q+2 d=21, b_{1} q+b_{1} q+d=18$.

Since $b_{3}=b_{1} q^{2}=b_{1} q+2 d$, we have $\left\{\begin{array}{l}b_{1} q^{2}=b_{1} q+d, \\ b_{1}+b_{1} q+2 d=21, \\ b_{1} q+b_{1} q+d=18,\end{array} \Rightarrow\right.$

Substituting this value of $d$ into the second and third equations, we get

$$
\begin{aligned}
& \Rightarrow\left\{\begin{array}{l}
b_{1}+b_{1} q+2\left(b_{1} q^{2}-b_{1} q\right)=21, \\
2 b_{1} q+b_{1} q^{2}-b_{1} q=18,
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ b _ { 1 } + 2 b _ { 1 } q ^ { 2 } - b _ { 1 } q = 2 1 , } \\
{ b _ { 1 } q + b _ { 1 } q ^ { 2 } = 1 8 , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}\left(2 q^{2}-q+1\right)=21, \\
b_{1}\left(q+q^{2}\right)=18,
\end{array} \Leftrightarrow\right.\right.
\end{aligned}
$$

$$
\Leftrightarrow\left\{\begin{array} { l } 
{ b _ { 1 } = \frac { 2 1 } { 2 q ^ { 2 } - q + 1 } , } \\
{ \frac { 2 1 ( q + q ^ { 2 } ) } { 2 q ^ { 2 } - q + 1 } = 1 8 , }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ b _ { 1 } = \frac { 2 1 } { 2 q ^ { 2 } - q + 1 } , } \\
{ 5 q ^ { 2 } - 1 3 q + 6 = 0 , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}=\frac{21}{2 q^{2}-q+1} \\
{\left[\begin{array}{l}
q_{1}=2 \\
q_{2}=\frac{3}{5}
\end{array}\right.}
\end{array}\right.\right.\right.
$$

From this:

1) $\left\{\begin{array}{l}b_{1}^{\prime}=\frac{21}{8-2+1}=3, \\ q_{1}=2,\end{array}\right.$ or 2$)\left\{\begin{array}{l}b_{1}^{\prime \prime}=\frac{21}{2 \cdot \frac{9}{25}-\frac{3}{5}+1}=18.75, \\ q_{2}=\frac{3}{5} .\end{array}\right.$

Then $q_{1}=2, \mathrm{~d}_{1}=3 \cdot 4-3 \cdot 2=6, b_{1}^{\prime}=3, b_{2}^{\prime}=3 \cdot 2=6, b_{3}^{\prime}=3 \cdot 4=12$, $b_{4}^{\prime}=6+12=18, \quad q_{2}=\frac{3}{5}, \quad d_{2}=18.75 \cdot \frac{9}{25}-18.75 \cdot \frac{3}{5}=-4.5, \quad b_{1}^{\prime \prime}=18.75$, $b_{2}^{\prime \prime}=11.25, b_{3}^{\prime \prime}=6.75, b_{4}^{\prime \prime}=2.25$.

Answer: 1$) 3,6,12,18 ; 2) 18.75 ; 11.25 ; 6.75 ; 2.25$.