Đáp án:
$\begin{array}{l}
4{S_n} = {S_{2n}}\\
\Rightarrow 4.\frac{{\left( {2{u_1} + \left( {n - 1} \right)d} \right).n}}{2} = n.\left( {2{u_1} + \left( {2n - 1} \right).d} \right)\\
\Rightarrow 4{u_1} + 2\left( {n - 1} \right)d = 2{u_1} + \left( {2n - 1} \right)d\\
\Rightarrow 2{u_1} = d\\
{u_5} = 18 \Rightarrow {u_1} + 4d = 18\\
\Rightarrow \left\{ \begin{array}{l}
{u_1} + 4d = 18\\
2{u_1} = d
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{u_1} = 2\\
d = 4
\end{array} \right.
\end{array}$