Cho cấp số cộng \(\left( {{u_n}} \right)\) thỏa \(\left\{ \begin{array}{l}{u_1} + {u_2} + {u_3} = 27\\u_1^2 + u_2^2 + u_3^2 = 275\end{array} \right.\). Tính \({u_2}.\)

Ta có \(\left\{ \begin{array}{l}{u_1} + {u_2} + {u_3} = 27\\u_1^2 + u_2^2 + u_3^2 = 275\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{u_1} + \left( {{u_1} + d} \right) + \left( {{u_1} + 2d} \right) = 27\\u_1^2 + {\left( {{u_1} + d} \right)^2} + {\left( {{u_1} + 2d} \right)^2} = 275\end{array} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{u_1} + d = 9}&{\left( 1 \right)}\\{u_1^2 + {{\left( {{u_1} + d} \right)}^2} + {{\left( {{u_1} + 2d} \right)}^2} = 275}&{\left( 2 \right)}\end{array}} \right..\)

Từ \(\left( 1 \right)\) suy ra \(d = 9 - {u_1}\). Thay vào \(\left( 2 \right)\), ta được

\(u_1^2 + {\left( {{u_1} + 9 - {u_1}} \right)^2} + {\left[ {{u_1} + 2\left( {9 - {u_1}} \right)} \right]^2} = 275\)\( \Leftrightarrow u_1^2 - 18{u_1} + 65 = 0\) \( \Leftrightarrow {u_1} = 13\) hoặc \({u_1} = 5\).

Vậy \(\left\{ \begin{array}{l}{u_1} = 13\\d =  - 4\end{array} \right.\) hoặc \(\left\{ \begin{array}{l}{u_1} = 5\\d = 4\end{array} \right. \Rightarrow {u_2} = {u_1} + d = 9\)