Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 182

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\[1)\ a_{6} = 20;\ \ S_{6} = 102:\]

\[a_{6} = a_{1} + 5d = 20.\]

\[S_{6} = \frac{2a_{1} + 5d}{2} \cdot 6 =\]

\[= \left( 2a_{1} + 5d \right) \cdot 3\]

\[\left( 2a_{1} + 5d \right) \cdot 3 = 102\]

\[2a_{1} + 5d = 34\]

\[\left\{ \begin{matrix} a_{1} + 5d = 20\ \ \ \\ 2a_{1} + 5d = 34 \\ \end{matrix} \right.\ ( - )\]

\[- a_{1} = - 14\]

\[a_{1} = 14.\]

\[5d = 20 - a_{1} = 20 - 14 = 6\]

\[5d = 6\]

\[d = 1,2.\]

\[Ответ:a_{1} = 14;\ \ d = 1,2.\]

\[2)\ a_{7} = 9;\ \ S_{7} = 98:\]

\[a_{7} = a_{1} + 6d = 9.\]

\[S_{7} = \frac{2a_{1} + 6d}{2} \cdot 7 = 98\ \ \ | \cdot \frac{2}{7}\]

\[2a_{1} + 6d = 28.\]

\[\left\{ \begin{matrix} a_{1} + 6d = 9\ \ \ \ \\ 2a_{1} + 6d = 28 \\ \end{matrix} \right.\ \ ( - )\]

\[- a_{1} = - 19\]

\[a_{1} = 19.\]

\[6d = 9 - a_{1} = 9 - 19 = - 10\]

\[d = - \frac{10}{6} = - \frac{5}{3} = - 1\frac{2}{3}.\]

\[Ответ:a_{1} = 19;d = - 1\frac{2}{3}.\]