Cho cấp số cộng \(\left( {{u_n}} \right)\) thỏa mãn \(\left\{ \begin{array}{l}{u_1} + {u_7} = 26\\{u_2}^2 + {u_6}^2 = 466\end{array} \right..\) Mệnh đề nào sau đây đúng?

Ta có

\(\left\{ \begin{array}{l}{u_1} + {u_7} = 26\\{u_2}^2 + {u_6}^2 = 466\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}2{u_1} + 6d = 26\\{\left( {{u_1} + d} \right)^2} + {\left( {{u_1} + 5d} \right)^2} = 466\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}{u_1} = 13 - 3d\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\{\left( {{u_1} + d} \right)^2} + {\left( {{u_1} + 5d} \right)^2} = 466\,\,\left( 2 \right)\end{array} \right..\)

Thay (1) và (2) ta được: \({\left( {13 - 2d} \right)^2} + {\left( {13 + 2d} \right)^2} = 466 \Leftrightarrow 8{d^2} + 338 = 466\)

\( \Leftrightarrow \left[ \begin{array}{l}d = 4 \Rightarrow {u_1} = 1\\d =  - 4 \Rightarrow {u_1} = 25\end{array} \right.\)