Решебник по алгебре 9 класс Мерзляк Задание 694

 

\[\boxed{\mathbf{694\ (694).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ a_{n} = n + 4\]

\[a_{1} = 1 + 4 = 5\]

\[a_{2} = 2 + 4 = 6\]

\[a_{3} = 3 + 4 = 7\]

\[a_{4} = 4 + 4 = 8\]

\[2)\ a_{n} = 4n - 3\]

\[a_{1} = 4 - 3 = 1\]

\[a_{2} = 8 - 3 = 5\]

\[a_{3} = 12 - 3 = 9\]

\[a_{4} = 16 - 3 = 13\]

\[3)\ a_{n} = \frac{n}{n² + 1}\]

\[a_{1} = \frac{1}{2}\]

\[a_{2} = \frac{2}{4 + 1} = \frac{2}{5}\]

\[a_{3} = \frac{3}{9 + 1} = \frac{3}{10}\]

\[a_{4} = \frac{4}{16 + 1} = \frac{4}{17}\]

\[4)\ a_{n} = \frac{2^{n}}{n}\]

\[a_{1} = \frac{2}{1} = 2\]

\[a_{2} = \frac{4}{2} = 2\]

\[a_{3} = \frac{8}{3}\]

\[a_{4} = \frac{16}{4} = 4\]

\[\boxed{\mathbf{694.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[Пусть\ числа\ a_{1},\ a_{2},\ a_{3}.\]

\[Получаем:\]

\[a_{1} + a_{2} + a_{3} = 21\ \]

\[a_{1} + a_{1} + d + a_{1} + 2d = 21\ \ \]

\[3a_{1} + 3d = 21\ \ \ |\ :3,\ \ \]

\[a_{1} + d = 7\ \]

\[\ d = 7 - a_{1}.\]

\[Значит,\ образуют\ \]

\[геометрическую\ \]

\[прогрессию\ числа\ \]

\[a_{1} + 2,a_{2} + 3,a_{3} + 9.\]

\[Отсюда:\]

\[\ \left( a_{2} + 3 \right)^{2} = \left( a_{1} + 2 \right)\left( a_{3} + 9 \right)\]

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

\[= \left( a_{1} + 2 \right)\left( a_{1} + 2d + 9 \right)\]

\[(7 + 3)^{2} =\]

\[= \left( a_{1} + 2 \right)\left( a_{1} + 2 \cdot \left( 7 - a_{1} \right) + 9 \right),\]

\[10^{2} =\]

\[= \left( a_{1} + 2 \right)\left( a_{1} + 14 - 2a_{1} + 9 \right)\ \]

\[10^{2} = \left( a_{1} + 2 \right)\left( 23 - a_{1} \right),\]

\[100 = 23a_{1} - a_{1}^{2} + 46 - 2a_{1}\text{\ \ }\]

\[a_{1}^{2} - 21a_{1} + 54 = 0\]

\[a_{1} + a_{2} = 21;\ \ a_{1} = 18\]

\[a_{1}a_{2} = 54;\ \ a_{1} = 3\]

\[при\ \ a_{1} = 3 \Longrightarrow \ \ d = 7 - 3 = 4:\]

\[a_{2} = 3 + 4 = 7;\ \]

\[a_{3} = a_{2} + d = 7 + 4 = 11;\]

\[при\ a_{1} = 18 \Longrightarrow d =\]

\[= 7 - 18 = - 11:\]

\[a_{2} = 18 - 11 = 7;\ \]

\[a_{3} = 7 - 11 =\]

\[= - 4 < 0 \Longrightarrow не\]

\[\ удовлетворяет.\]

\[Ответ:3;7;11.\]