Đáp án:
$\begin{array}{l}
{S_{20}} = \dfrac{{\left( {2.{u_1} + 19d} \right).20}}{2} = 20{u_1} + 38d\\
{s_{10}} = \dfrac{{\left( {2{u_1} + 9d} \right).10}}{2} = 10{u_1} + 45d\\
{S_{15}} = \dfrac{{\left( {2{u_1} + 14d} \right).15}}{2} = 15{u_1} + 105d\\
{S_5} = \dfrac{{\left( {2{u_1} + 4d} \right).5}}{2} = 5{u_1} + 10d\\
\Rightarrow \left\{ \begin{array}{l}
20{u_1} + 38d = 2.\left( {10{u_1} + 45d} \right)\\
15{u_1} + 105d = 3.\left( {5{u_1} + 10d} \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
d = 0\\
d = 0
\end{array} \right.
\end{array}$
Vậy ko có cấp số cộng thỏa mãn yêu cầu
