Tìm cấp số cộng \({a_1},{a_2},{a_3},{a_4},{a_5}\) biết rằng: \({a_1} + {a_3} + {a_5} = -12;{\rm{ }}{a_1}{a_3}{a_5} = 80\)

Gọi công sai là d, ta có

\(\begin{array}{l}\left\{ \begin{array}{l}{a_1} + {a_3} + {a_5} =  - 12\\{a_1}{a_3}{a_5} = 80\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{a_1} + {a_1} + 2d + {a_1} + 4d =  - 12\\{a_1}\left( {{a_1} + 2d} \right)\left( {{a_1} + 4d} \right) = 80\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}3{a_1} + 6d =  - 12\\{a_1}\left( {{a_1} + 2d} \right)\left( {{a_1} + 4d} \right) = 80\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{a_1} + 2d =  - 4\\{a_1}\left( {{a_1} + 2d} \right)\left( {{a_1} + 4d} \right) = 80\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{a_1} =  - 4 - 2d\\\left( { - 4 - 2d} \right)\left( { - 4 - 2d + 2d} \right)\left( { - 4 - 2d + 4d} \right) = 80\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{a_1} =  - 4 - 2d\\\left( { - 4 - 2d} \right)\left( { - 4} \right)\left( { - 4 + 2d} \right) = 80\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{a_1} =  - 4 - 2d\\ - 4\left( {16 - 4{d^2}} \right) = 80\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{a_1} =  - 4 - 2d\\16 - 4{d^2} =  - 20\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{a_1} =  - 4 - 2d\\{d^2} = 9\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{a_1} =  - 4 - 2d\\d =  \pm 3\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}d = 3,{a_1} =  - 10\\d =  - 3,{a_1} = 2\end{array} \right.\end{array}\)

Các cấp số cộng phải tìm là

2, -1, -4, -7, -10 Và -10, -7, -4, -1, 2