Bài tập 3.21 trang 124 SBT Toán 11

\(\begin{array}{l}
\left\{ \begin{array}{l}
{a_2} + {a_4} + ... + {a_{2n}} = 126\\
{a_2} + {a_{2n}} = 42
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{a_1} + d + {a_1} + 3d + ... + {a_1} + \left( {2n - 1} \right)d = 126\\
{a_1} + d + {a_1} + \left( {2n - 1} \right)d = 42
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
n{a_1} + \left[ {1 + 3 + 5 + ... + \left( {2n - 1} \right)} \right]d = 126\\
{a_1} + nd = 21
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
n{a_1} + \frac{{n\left( {1 + 2n - 1} \right)}}{2}.d = 126\\
{a_1} + nd = 21
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
n{a_1} + {n^2}d = 126\\
{a_1} + nd = 21
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
n.21 = 126\\
{a_1} + nd = 21
\end{array} \right. \Leftrightarrow n = 6
\end{array}\)

-- Mod Toán 11 HỌC247