Решебник по алгебре 9 класс Мерзляк Задание 695

 

\[\boxed{\mathbf{695\ (695).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ b_{n} = \frac{10}{n}\]

\[b_{2} = \frac{10}{2} = 5,\ \ b_{7} = \frac{10}{7},\]

\[\text{\ \ }b_{100} = \frac{10}{100} = 0,1\]

\[2)\text{\ b}_{n} = 5 - 2n\]

\[b_{2} = 5 - 4 = 1,\ \ \]

\[b_{7} = 5 - 14 = - 9,\ \ \]

\[b_{100} = 5 - 200 = - 195\]

\[\ 3)\ b_{n} = n² + 2n\]

\[b_{2} = 4 + 4 = 8,\ \ \]

\[b_{7} = 49 + 14 = 63,\ \ \]

\[b_{100} = 10\ 000 + 200 = 10\ 200\]

\[4)\ b_{n} = ( - 1)^{n + 1}\]

\[b_{2} = ( - 1)^{3} = - 1,\ \ \]

\[b_{7} = ( - 1)^{8} = 1,\]

\[\text{\ \ }b_{100} = ( - 1)^{101} = - 1\]

\[\boxed{\mathbf{695.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[Арифметическая\ прогрессия:\]

\[\ a_{1} + a_{2} + a_{3} = 30.\ \ \]

\[a_{1} + a_{1} + d + a_{1} + 2d = 30\ \]

\[\ 3a_{1} + 3d = 30\ \ |\ :3\ \]

\[a_{1} + d = 10\]

\[d = 10 - a_{1}.\]

\[Геометрическая\ прогрессия:\ \]

\[a_{1} - 5,\ a_{2} - 4,\ a_{3}\text{.\ }\]

\[\left( a_{2} - 4 \right)^{2} = a_{3}\left( a_{1} - 5 \right)\text{\ \ \ }\]

\[\left( a_{1} + d - 4 \right)^{2} =\]

\[= (a_{1} + 2d)(a_{1} - 5)\]

\[(10 - 4)^{2} =\]

\[= \left( a_{1} + 2 \cdot \left( 10 - a_{1} \right) \right) \cdot \left( a_{1} - 5 \right)\]

\[6^{2} = \left( a_{1} + 20 - 2a_{1} \right)\left( a_{1} - 5 \right)\]

\[36 = \left( - a_{1} + 20 \right)\left( a_{1} - 5 \right)\]

\[36 = - a_{1}^{2} + 5a_{1} + 20a_{1} - 100\]

\[a_{1}^{2} - 25a_{1} + 136 = 0\]

\[a_{1} + a_{2} = 25,\ \ a_{1} = 8\]

\[a_{1}a_{2} = 136,\ \ a_{1} = 17.\]

\[при\ a_{1} = 8 \Longrightarrow \ d = 10 - 8 = 2:\ \ \]

\[a_{2} = 8 + 2 = 10,\ \ \]

\[a_{3} = 10 + 2 = 12;\]

\[при\ a_{1} = 17 \Longrightarrow \ \ d =\]

\[= 10 - 17 = - 7:\ \ \]

\[a_{2} = 17 - 7 = 10,\ \ \]

\[a_{3} = 10 - 7 = 3.\]

\[Ответ:8,\ 10,\ 12\ \ \ или\ \ 17,\ 10,\ 3.\]