Решебник по алгебре 9 класс Макарычев Задание 579

 

\[\boxed{\text{579}\text{\ (579)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{1}{3};\ - 1;\ldots\]

\[a_{1} = \frac{1}{3};\ \ a_{2} = - 1;\ \ \]

\[d = a_{2} - a_{1} = - 1 - \frac{1}{3} = - \frac{4}{3}\]

\[a_{n} = \frac{1}{3} - \frac{4}{3} \cdot (n - 1)\]

\[a_{10} = \frac{1}{3} - \frac{4}{3} \cdot 9 = \frac{1}{3} - 12 =\]

\[= - 11\frac{2}{3}.\]

\[\textbf{б)}\ 2,3;\ \ 1;\ldots\ \]

\[a_{1} = 2,3;\ \ \ \ a_{2} = 1;\ \ \]

\[d = a_{2} - a_{1} = 1 - 2,3 = - 1,3;\]

\[a_{n} = 2,3 - 1,3 \cdot (n - 1);\]

\[a_{10} = 2,3 - 1,3 \cdot (10 - 1) =\]

\[= - 9,4.\]

\[\boxed{\text{579.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[b_{1} = 4,2;\ \ \ b_{10} = b_{1} + 9d = 15,9:\ \ \ \ \]

\[4,2 + 9d = 15,9\]

\[9d = 11,7\]

\[d = 1,3\]

\[S_{15} = \frac{2c_{1} + d(n - 1)}{2} \cdot n =\]

\[= \frac{8,4 + 1,3 \cdot 14}{2} \cdot 15 = 199,5.\]