\[{u_{n + 1}} = {u_n} + 3 \Rightarrow \left( {{u_n}} \right)\] là CSC có công sai\[d = 3.\]

\[{u_3} = {u_1} + 2d \Rightarrow {u_1} = {u_3} - 2d = - 2 - 2.3 = - 8\]

Vậy số hạng tổng quát của CSC trên là

\[{u_n} = {u_1} + \left( {n - 1} \right)d = - 8 + \left( {n - 1} \right).3 = 3n - 11.\]

\(\left\{ {\begin{array}{*{20}{c}}{{u_2} = 2017}\\{{u_5} = 1945}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{u_1} + d = 2017}\\{{u_1} + 4d = 1945}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{u_1} = 2041}\\{d = - 24}\end{array}} \right.\)

\[ \Rightarrow {u_{2018}} = {u_1} + 2017d\]

\[ = 2041 + 2017( - 24) = - 46367\]

\(\left\{ {\begin{array}{*{20}{c}}{{u_3} + {u_5} = 5}\\{{u_3}.{u_5} = 6}\end{array}} \right. \Rightarrow {u_3},{u_5}\) là nghiệm của phương trình

\[{X^2} - 5X + 6 = 0 \Rightarrow \left[ {\begin{array}{*{20}{c}}{X = 3}\\{X = 2}\end{array}} \right. \Rightarrow \left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{{u_3} = 3}\\{{u_5} = 2}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{{u_3} = 2}\\{{u_5} = 3}\end{array}} \right.}\end{array}} \right.\]

TH1 : \(\left\{ {\begin{array}{*{20}{c}}{{u_3} = 3}\\{{u_5} = 2}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{u_1} + 2d = 3}\\{{u_1} + 4d = 2}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{u_1} = 4}\\{d = - \frac{1}{2}}\end{array}} \right.\)

TH2 : \(\left\{ {\begin{array}{*{20}{c}}{{u_3} = 2}\\{{u_5} = 3}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{u_1} + 2d = 2}\\{{u_1} + 4d = 3}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{u_1} = 1}\\{d = \frac{1}{2}}\end{array}} \right.\)

Vậy\(\left[ {\begin{array}{*{20}{c}}{{u_1} = 1}\\{{u_1} = 4}\end{array}} \right.\)

Đáp án cần chọn là: A