Найдите сумму шестнадцати первых членов арифметической прогрессии (an), если a5+a7-a12=-9 и a3+a20=74.

Ответ:

\[\left\{ \begin{matrix} a_{5} + a_{7} - a_{12} = - 9 \\ a_{3} + a_{20} = 74\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} a_{1} - d = - 9\ \ \ \ \ | \cdot 12 \\ 2a_{1} + 21d = 74\ \ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ } \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2a_{1} - 2d = - 18 \\ 2a_{1} + 21d = 74 \\ \end{matrix}\ \ \ ( - ) \right.\ \]

\[\left\{ \begin{matrix} - 23d = - 92 \\ a_{1} = - 9 + d \\ \end{matrix}\text{\ \ \ \ \ \ } \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} d = 4\ \ \ \ \ \\ a_{1} = - 5 \\ \end{matrix} \right.\ \]

\[S_{16} = \left( 2a_{1} + 15d \right) \cdot 8 =\]

\[= ( - 10 + 60) \cdot 8 = 50 \cdot 8 =\]

\[= 400.\]

\[Ответ:400.\]