Решебник по алгебре 9 класс Макарычев Задание 692

 

\[\boxed{\text{692}\text{\ (692)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ a_{1} = 1,\ \ d = 1,\]

\[a_{n} = a_{1} + d(n - 1) \Longrightarrow 1 + n - 1 = n \Longrightarrow\]

\[S_{n} = \frac{a_{1} + a_{n}}{2} \cdot n = \frac{1 + n}{2} \cdot n;\]

\[5a_{n + 1} = S_{n}\]

\[5(n + 1) = \frac{1 + n}{2} \cdot n\]

\[n = 10\]

\[a_{n + 1} = n + 1 = 11.\]

\[\textbf{б)}\ a_{1} = 1,\ \ d = 1,\]

\[a_{n} = a_{1} + d(n - 1) = 1 + n - 1 = n,\]

\[S_{n} = \frac{a_{1} + a_{n}}{2} \cdot n = \frac{1 + n}{2} \cdot n\]

\[a_{n + 1} = S_{n}\]

\[n + 1 = \frac{1 + n}{2} \cdot n\]

\[n = 2 \Longrightarrow x_{n + 1} = n + 1 = 3.\]

\[\boxed{\text{692.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[= \frac{1000}{27} + \frac{7}{3} + 3 + 6 =\]

\[= \frac{1000 + 63 + 81 + 162}{27} =\]

\[= \frac{1306}{27} = 48\frac{10}{27}.\]

\[= \left( \frac{3}{2} \right)^{2} - 9 + 1 \cdot \frac{1}{8} - 4^{2} \cdot 16 =\]

\[= \frac{9}{4} - 9 + \frac{1}{8} - 256 = \frac{19}{8} - 265 =\]

\[= - 262\frac{5}{8}.\]