4.008. Find four numbers that form a geometric progression, where the sum of the extreme terms is -49, and the sum of the middle terms is 14.
## Solution.

From the condition, we have $\left\{\begin{array}{l}b_{1}+b_{4}=-49, \\ b_{2}+b_{3}=14 .\end{array}\right.$

Using formula (4.6), we get

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ b _ { 1 } + b _ { 1 } q ^ { 3 } = - 4 9 , } \\
{ b _ { 1 } q + b _ { 1 } q ^ { 2 } = 1 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}\left(1+q^{3}\right)=-49, \\
b_{1} q(1+q)=14
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
b_{1}(1+q)\left(1-q+q^{2}\right)=-49 \\
b_{1} q(1+q)=14
\end{array}\right.
\end{aligned}
$$

Dividing the first equation of the system by the second, we get

$$
\begin{aligned}
& \frac{b_{1}(1+q)\left(1-q+q^{2}\right)}{b_{1} q(1+q)}=-\frac{49}{14}, \quad \frac{q^{2}-q+1}{q}+\frac{49}{14}=0, \quad 14 q^{2}+35 q+14=0 \\
& 2 q^{2}+5 q+2=0 \text { i.e. } q'=-2, q''=-\frac{1}{2}, b_{1}'=7, b_{1}''=-56
\end{aligned}
$$

Then we get:

1) $b_{1}'=7, b_{2}'=-14, b_{3}'=28, b_{4}'=-56$;
2) $b_{1}''=-56, b_{2}''=28, b_{3}''=-14, b_{4}''=7$.

Answer: 1) 7, -14, 28, -56; 2) $-56,28,-14,7$.