ГДЗ по алгебре 9 класс Макарычев Задание 680

\[\boxed{\text{680}\text{\ (680)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ a_{1} = - 10\frac{1}{2},\ \ a_{2} = - 10\frac{1}{4},\ \ \]

\[d = a_{2} - a_{1} = - 10\frac{1}{4} + 10\frac{1}{2} = - \frac{41}{4} + \frac{21}{2} = \frac{1}{4},\]

\[a_{n} = a_{1} + d(n - 1) > 0 \Longrightarrow\]

\[- 10\frac{1}{2} + \frac{1}{4}n - \frac{1}{4} > 0\]

\[\frac{1}{4}n > 10\frac{3}{4}\]

\[n > 43 \Longrightarrow n = 44,\]

\[a_{44} = - 10\frac{1}{2} + \frac{1}{4} \cdot (44 - 1) = - \frac{42}{4} + \frac{43}{4} = \frac{1}{4}.\]

\[\textbf{б)}\ a_{1} = 8\frac{1}{2},\ \ a_{2} = 8\frac{1}{3},\ \]

\[d = a_{2} - a_{1} = 8\frac{1}{3} - 8\frac{1}{2} = \frac{25}{3} - \frac{17}{2} = - \frac{1}{6},\]

\[a_{n} = \frac{25}{3} - \frac{1}{6} \cdot (n - 1) \Longrightarrow\]

\[\frac{25}{3} - \frac{n}{6} + \frac{1}{6} < 0\]

\[n > 50 + 1 \Longrightarrow n > 51 \Longrightarrow n = 52,\]

\[a_{52} = 8,5 - \frac{1}{6} \cdot (53 - 1) = - \frac{1}{6}.\]