Решебник по алгебре 9 класс Макарычев Задание 576

\[\boxed{\text{576}\text{\ (576)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[\textbf{а)}\ b_{7} = b_{1} + d(7 - 1) = b_{1} + 6d;\]

\[\textbf{б)}\ b_{26} = b_{1} + d(26 - 1) =\]

\[= b_{1} + 25d;\]

\[\textbf{в)}\ b_{231} = b_{1} + d(231 - 1) =\]

\[= b_{1} + 230d;\]

\[\textbf{г)}\ b_{k} = b_{1} + d \cdot (k - 1);\]

\[\textbf{д)}\ b_{k + 5} = b_{1} + d \cdot (k + 5 - 1) =\]

\[= b_{1} + d(k + 4);\]

\[\textbf{е)}\ b_{2k} = b_{1} + d \cdot (2k - 1)\ \]

\[\boxed{\text{576.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[x_{1} = 10,\ \ d = 3:\ \ \]

\[x_{15} = x_{1} + d \cdot (15 - 1) =\]

\[= 10 + 3 \cdot 14 = 52;\]

\[x_{30} = 10 + 3 \cdot (30 - 1) = 97;\ \ \ \]

\[S_{n} = \frac{(x_{1} + x_{n})}{2} \cdot n;\]

\[S_{16} = \frac{x_{15} + x_{30}}{2} \cdot 16 =\]

\[= (52 + 97) \cdot 8 = 1192.\]