Theo đề bài ta có:
$\left \{ {{u_1+u_2+u_3+u_4=-20} \atop {u_1^2+u_2^2+u_3^2+u_4^2=280}} \right.$
⇔ $\left \{ {{4u_1+6d=-20} \atop {u_1^2+(u_1+d)^2+(u_1+2d)^2+(u_1+3d)^2=280}} \right.$
⇔ $\left \{ {{2u_1+3d=-10} \atop {4u_1^2+12u_1d+14d^2=280}} \right.$
⇔ $\left \{ {{2u_1=-3d-10} \atop {(3d+10)^2+6d(-3d-10)+14d^2=280}} \right.$
⇔ $\left \{ {{2u_1=-3d-10} \atop {5d^2=180}} \right.$
⇔ $\left \{ {{2u_1=-3d-10} \atop {d=±6}} \right.$
⇔ \(\left[ \begin{array}{l}\left \{ {{u_1=-14} \atop {d=6}} \right.\\\left \{ {{u_1=4} \atop {d=-6}} \right.\end{array} \right.\)
Vậy có hai cấp số cộng thỏa mãn đề bài là:
$-14;-8;-2;4$ và $4;-2;-8;-14$
