Đáp án:
Giải thích các bước giải:
Có:
\(\begin{array}{l}
\frac{{{S_{20}}}}{5} = \frac{{20(2{U_1} + 19d)}}{{2.5}}\\
\frac{{{S_{10}}}}{3} = \frac{{10(2{U_1} + 9d)}}{{2.3}}\\
\frac{{{S_5}}}{2} = \frac{{5.(2{U_1} + 4d)}}{{2.2}}
\end{array}\)
Đề⇒ \(\begin{array}{l}
\left\{ \begin{array}{l}
2(2{U_1} + 19d) = \frac{{5(2{U_1} + 9d)}}{3}\\
2(2{U_1} + 19d) = \frac{{5.(2{U_1} + 4d)}}{4}
\end{array} \right. \to \left\{ \begin{array}{l}
4{U_1} + 38d = \frac{{10}}{3}{U_1} + 15d\\
4{U_1} + 38d = \frac{5}{2}{U_1} + 5d
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\frac{2}{3}{U_1} + 23d = 0\\
\frac{3}{2}{U_1} + 33d = 0
\end{array} \right. \to \left\{ \begin{array}{l}
{U_1} = \frac{{ - 69}}{2}d\\
\frac{3}{2}.\frac{{ - 69}}{2}d + 33d = 0
\end{array} \right. \to \left\{ \begin{array}{l}
d = 0\\
{U_1} = 0
\end{array} \right.
\end{array}\)
