Решебник по алгебре 9 класс Макарычев Задание 678

 

\[\boxed{\text{678}\text{\ (678)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[- 2,94 = 1,26 - 0,3 \cdot (n - 1)\]

\[0,3n = 1,26 + 2,94 + 0,3\]

\[0,3n = 4,5\]

\[n = 15.\]

\[\textbf{б)}\ a_{5} = a_{1} + 4d\]

\[a_{1} = a_{5} - 4d = - 3,7 - 4 \cdot ( - 0,6) = - 1,3;\]

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

\[- 9,7 = - 1,3 - 0,6 \cdot (n - 1)\]

\[0,6n = 9,7 - 1,3 + 0,6\]

\[0,6n = 9\]

\[n = 15.\]

\[\boxed{\text{678.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[S_{n} = \frac{3}{4} \cdot \left( 5^{n} - 1 \right),\ \ \]

\[x_{1} = S_{1} = 3,\ \ \]

\[x_{2} = S_{2} - x_{1} = \frac{3}{4} \cdot 24 - 3 = 15,\]

\[x_{2} = x_{1} \cdot q,\ \ q = \frac{x_{2}}{x_{1}} = 5,\]

\[S_{n} = x_{1} \cdot \frac{q^{n} - 1}{q - 1} = 3 \cdot \frac{5^{n} - 1}{4} =\]

\[= \frac{3}{4} \cdot \left( 5^{n} - 1 \right) \Longrightarrow\]

\[\Longrightarrow геометрическая\]

\[прогрессия.\]