Решебник по алгебре 9 класс Макарычев Задание 657

 

\[\boxed{\text{657}\text{\ (657)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[b_{1} > 0,\ \ b_{2} > 0,\ \ldots,\ b_{n} > 0\ \]

\[b_{1} + b_{2} = b_{1} + b_{1} \cdot q = 8\]

\[b_{3} + b_{4} = b_{1} \cdot q^{2} + b_{1} \cdot q^{3} = 72;\]

\[\left\{ \begin{matrix} b_{1}(1 + q) = 8\ \ \ \ \ \ \ \ \ \ \\ b_{1} \cdot q^{2} \cdot (1 + q) = 72 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} q^{2} = 9\ \ \ \ \ \ \ \\ b_{1} = \frac{8}{q + 1} \\ \end{matrix} \Longrightarrow \right.\ \left\{ \begin{matrix} q = 3\ \ \\ b_{1} = 2 \\ \end{matrix} \right.\ ,\ \ \]

\[q = - 3 \Longrightarrow не\ подходит\ по\ условию.\]

\[S_{n} = \frac{b_{1} \cdot \left( q^{n} - 1 \right)}{q - 1} = 2 \cdot \frac{3^{n} - 1}{2} = 3^{n} - 1 = 242 \Longrightarrow\]

\[\Longrightarrow 3^{n} = 243 \Longrightarrow n = 5.\]

\[Ответ:\ \ 5\ членов\ прогрессии\ нужно\ сложить.\]

\[\boxed{\text{657.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ a_{1} = 2,\ \ d = 2,\]

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

\[= 2 + 2n - 2 = 2n\]

\[2n = 200 \Longrightarrow n = 100;\]

\[S_{100} = \frac{a_{1} + a_{100}}{2} \cdot 100 =\]

\[= (2 + 200) \cdot 50 = 10100.\]

\[\textbf{б)}\ a_{1} = 1,\ \ d = 2,\ \ \]

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

\[= 1 + 2n - 2 = 2n - 1\]

\[2n - 1 = 149\]

\[2n = 150 \Longrightarrow n = 75,\]

\[S_{75} = \frac{a_{1} + a_{75}}{2} \cdot 75 =\]

\[= \frac{1 + 149}{2} \cdot 75 = 5625.\]

\[\textbf{в)}\ a_{1} = 102,\ \ d = 3,\]

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

\[= 102 + 3n - 3 = 3n + 99 =\]

\[= 198\]

\[3n + 99 = 198\]

\[3n = 99\]

\[n = 33,\]

\[S_{33} = \frac{a_{1} + a_{33}}{2} \cdot 33 =\]

\[= \frac{102 + 198}{2} \cdot 33 = 150 \cdot 33 =\]

\[= 4950.\]