Tính S100. Biết u9-u4=10 & u3.u6=55 Giúp mk vs ạ

Đáp án:

$\begin{array}{l}
{u_9} = {u_1} + 8d\\
{u_4} = {u_1} + 3d\\
{u_3} = {u_1} + 2d\\
{u_6} = {u_1} + 5d\\
 \Rightarrow \left\{ \begin{array}{l}
{u_1} + 8d – {u_1} – 3d = 10\\
\left( {{u_1} + 2d} \right)\left( {{u_1} + 5d} \right) = 55
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
5d = 10\\
\left( {{u_1} + 2d} \right)\left( {{u_1} + 5d} \right) = 55
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
d = 2\\
\left( {{u_1} + 4} \right)\left( {{u_1} + 10} \right) = 55
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
d = 2\\
u_1^2 + 14{u_1} – 15 = 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
d = 2\\
\left( {{u_1} – 1} \right)\left( {{u_1} + 15} \right) = 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
d = 2\\
\left[ \begin{array}{l}
{u_1} = 1\\
{u_1} =  – 15
\end{array} \right.
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
{S_{100}} = \dfrac{{\left( {2{u_1} + 99d} \right).100}}{2} = \dfrac{{\left( {2.1 + 99.2} \right).100}}{2} = 10000\\
{S_{100}} = \dfrac{{\left( {2{u_1} + 99d} \right).100}}{2} = 8400
\end{array} \right.
\end{array}$