Cho cấp số cộng \(({u_n})\) có công sai \(d > 0\); \(\left\{ {\begin{array}{*{20}{c}}{{u_{31}} + {u_{34}} = 11}\\{{u^2}_{31} + {u^2}_{34} = 101}\end{array}} \right.\). Hãy tìm số hạng tổng quát của cấp số cộng đó.

Ta có:

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{{u_{31}} + {u_{34}} = 11}\\{{u^2}_{31} + {u^2}_{34} = 101}\end{array}} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{u_{34}} = 11 – {u_{31}}\\{u^2}_{31} + {(11 – {u_{31}})^2} = 101\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}{u_{31}} = 1\\{u_{34}} = 10\end{array} \right.\\\left\{ \begin{array}{l}{u_{31}} = 10\\{u_{34}} = 1\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}{u_1} =  – 89\\d = 3\end{array} \right.\\\left\{ \begin{array}{l}{u_1} = 100\\d =  – 3\end{array} \right.\end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}{u_1} =  – 89\\d = 3\end{array} \right.(d > 0)\end{array}\)

\({u_n} =  – 89 + (n – 1)3 = 3n – 92\)

Chọn C.