4.077 The sum of four numbers forming a geometric progression is -40, and the sum of their squares is 3280. Find this progression.
Solution. Let $b, b q, b q^{2}, b q^{3}$ be the given numbers: Then
$\left\{\begin{array}{l}b\left(1+q+q^{2}+q^{3}\right)=-40, \\ b^{2}\left(1+q^{2}+q^{4}+q^{6}\right)=3280\end{array} \Leftrightarrow\left\{\begin{array}{l}b(1+q)\left(1+q^{2}\right)=-40, \\ b^{2}\left(1+q^{2}\right)\left(1+q^{4}\right)=3280 .\end{array}\right.\right.$

Square the first equation and divide the result by the second equation:

$\frac{(1+q)^{2}\left(1+q^{2}\right)}{1+q^{4}}=\frac{20}{41} \Leftrightarrow \frac{\left(1+q^{2}+2 q\right)\left(1+q^{2}\right)}{\left(1+q^{2}\right)^{2}-2 q^{2}}=\frac{20}{41}$.

Let $p=1+q^{2}$, we have: $\frac{(p+2 q) p}{p^{2}-2 q^{2}}=\frac{20}{21} \Rightarrow 21 p^{2}+82 p q+40 q^{2}=0 \Rightarrow$

$\Rightarrow p=\frac{-41 q \pm \sqrt{(41 q)^{2}-840 q^{2}}}{21}=\frac{-41 q \pm 29 q}{21} \Rightarrow\left[\begin{array}{l}p=-\frac{10}{3} q, \\ p=-\frac{4}{7} q\end{array} \Rightarrow\right.$

$\Rightarrow\left\{\begin{array}{c}3 q^{2}+10 q+3=0, \\ 7 q^{2}+4 q+7=0\end{array} \Rightarrow\left[\begin{array}{c}q=-3, b=2, \\ q=-\frac{1}{3}, b=-54 .\end{array}\right.\right.$

Answer: $2,-6,18,-54$.