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МатематикаРешебник по алгебре 9 класс Макарычев Задание 611
Задание 611
\[\boxed{\text{611}\text{\ (611)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[x_{1} = 21,\ \ d = - 0,5:\ \text{\ \ \ \ \ \ \ \ \ \ }\]
\[x_{6} = 21 - 0,5 \cdot (6 - 1) = 18,5;\]
\[x_{25} = 21 - 0,5 \cdot (25 - 1) = 21 - 12 = 9;\ \]
\[S_{n} = \frac{\left( x_{1} + x_{n} \right)}{2} \cdot n;\]
\[S_{20} = \frac{18,5 + 9}{2} \cdot 20 = 27,5 \cdot 10 = 275.\]
\[\boxed{\text{611.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[x_{1} = a,\ \ x_{2} = a + d,\]
\[\text{\ \ }x_{3} = a + 2d,\]
\[\text{\ \ }x_{1} + 1 = a + 1,\]
\[x_{2} + 1 = (a + 1)q,\ \ \]
\[x_{3} + 4 = (a + 1)q^{2},\]
\[\left\{ \begin{matrix} a + d = (a + 1)q - 1\ \ \ \ \ \\ a + 2d = (a + 1)q^{2} - 4 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} d = (a + 1)(a - 1)\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ 2d + 3 = (a + 1)\left( q^{2} - 1 \right) \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 2d + 3 = \frac{d}{q - 1} \cdot \left( q^{2} - 1 \right) \\ q - 1 = \frac{d}{a + 1}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 2 + \frac{3}{d} = q + 1 \\ a + 1 = \frac{d}{q - 1} \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} q = \frac{3}{d} + 1\ \ \ \ \ \ \ \\ a = \frac{d}{q - 1} - 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} q = \frac{3 + d\ }{d}\text{\ \ } \\ a = \frac{d^{2}}{3} - 1 \\ \end{matrix} \right.\ \]
\[x_{1} + x_{2} + x_{3} = 3a + 3d = 15,\]
\[\ \ a + d = 5,\]
\[\frac{d^{2}}{3} - 1 + d = 15,\ \ \]
\[d^{2} + 3d - 3 = 15,\]
\[d^{2} + 3d - 18 = 0,\]
\[\text{\ \ }по\ теореме\ Виета:\ \ d_{1} = - 6,\ \ \]
\[d_{2} = 3,\]
\[1)\ d_{1} = - 6,\ \ a + d = 5,\ \ \]
\[x_{1} = a = 11,\]
\[x_{2} = 5,\ \ x_{3} = - 1 \Longrightarrow\]
\[\Longrightarrow не\ подходит,\ так\ как\ \]
\[по\ условию\ все\ числа\]
\[положительные;\]
\[2)\ d_{2} = 3,\ \ a + d = 5,\ \ \]
\[x_{1} = a = 2,\ \ x_{2} = 5,\ \ \]
\[x_{3} = 8.\]
\[Ответ:2;5;8.\]
