6.3 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.
6.3 Using the condition and formula (6.6), we obtain the system

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

From this, it follows that

$$
\frac{1-q+q^{2}}{q}=-\frac{49}{14}, \text { or } 2 q^{2}+5 q+2=0
$$

i.e., $q_{1}=-2, q_{2}=-0.5$. Then the four sought numbers are $7, -14, 28, -56$.

Answer: $7, -14, 28, -56$.