Cho cấp số cộng (un) thỏa mãn điều kiện:U1+ U2+ U3+ U4+-U5=20 

Đáp án đúng là: C

Theo đề bài ta có: \(\left\{ {\begin{array}{*{20}{c}}{{u_1} + {u_2} + {u_3} + {u_4} + {u_5} = 20}\\{u_1^2 + u_2^2 + u_3^2 + u_4^2 + u_5^2 = 170}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{u_1} + {u_1} + d + {u_1} + 2d + {u_1} + 3d + {u_1} + 4d = 20}\\{u_1^2 + {{\left( {{u_1} + d} \right)}^2} + {{\left( {{u_1} + 2d} \right)}^2} + {{\left( {{u_1} + 3d} \right)}^2} + {{\left( {{u_1} + 4d} \right)}^2} = 170}\end{array}} \right.\) 

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{5{u_1} + 10d = 20 \,\, (1)}\\{u_1^2 + {{\left( {{u_1} + d} \right)}^2} + {{\left( {{u_1} + 2d} \right)}^2} + {{\left( {{u_1} + 3d} \right)}^2} + {{\left( {{u_1} + 4d} \right)}^2} = 170\,\,(2)}\end{array}} \right.\)

Từ (1) suy ra: u₁ + 2d = 4 \(\Leftrightarrow \) u₁ = 4 − 2d thế vào (2) ta được:

\({(4 - 2d)^2} + {(4 - 2d + d)^2} + {(4 - 2d + 2d)^2} + {(4 - 2d + 3d)^2} + {\left( {4 - 2d + 4d} \right)^2} = 170\) \( \Leftrightarrow {(4 - 2\;{\rm{d}})^2} + {(4 - {\rm{d}})^2} + 16 + {(4 + {\rm{d}})^2} + {(4 + 2\;{\rm{d}})^2} = 170\)\( \Leftrightarrow 16 - 16\;{\rm{d}} + 4\;{{\rm{d}}^2} + 16 - 8\;{\rm{d}} + {{\rm{d}}^2} + 16 + 16 + 8\;{\rm{d}} + {{\rm{d}}^2} + 16 + 16\;{\rm{d}} + 4\;{{\rm{d}}^2} = 170\)

\( \Leftrightarrow 10\;{{\rm{d}}^2} + 80 = 170 \Leftrightarrow {{\rm{d}}^2} = 9\)\( \Leftrightarrow {\rm{d}} = \pm 3\).

•Với d = 3 \(\Rightarrow \) u₁ = 4 − 6 = −2.

•Với d = −3 \(\Rightarrow \) u₁ = 4 + 6 = 10.