ГДЗ по алгебре 9 класс Макарычев Задание 689

\[\boxed{\text{689}\text{\ (689)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ a_{1} = 10,\ \ a_{n} = 99,\ \ d = 1\]

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

\[99 = 10 + n - 1\]

\[99 = 9 + n\]

\[n = 90;\]

\[S_{90} = \frac{a_{1} + a_{90}}{2} \cdot 90 = (10 + 99) \cdot 45 = 4905.\]

\[\textbf{б)}\ a_{1} = 100,\ \ d = 1,\ \ a_{n} = a_{1} + d(n - 1),\]

\[999 = 100 + n - 1\]

\[999 = 99 + n\]

\[n = 900\]

\[S_{900} = \frac{a_{1} + a_{900}}{2} \cdot 900 = (100 + 999) \cdot 450 = 494\ 550.\]