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

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1. Đáp án:

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

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