Решебник по алгебре 9 класс Макарычев Задание 656

 

\[\boxed{\text{656}\text{\ (656)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[b_{2} = 6,\ \ b_{4} = 54,\ \ q > 0;\]

\[b_{4} = b_{2} \cdot q^{2}\]

\[q^{2} = \frac{b_{4}}{b_{2}}\]

\[q^{2} = \frac{54}{6} = 9\]

\[q = 3\]

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

\[\Longrightarrow b_{2} = b_{1} \cdot q \Longrightarrow b_{1} = \frac{b_{2}}{q} \Longrightarrow b_{1} = \frac{6}{3} = 2.\]

\[S_{7} = \frac{b_{1} \cdot (q^{7} - 1)}{q - 1} = \frac{2 \cdot (3^{7} - 1)}{3 - 1} = 2186.\]

\[\boxed{\text{656.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

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

\[\ \ d = 1\]

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

\[99 = 10 + n - 1\]

\[99 = 9 + n\]

\[n = 90;\]

\[S_{90} = \frac{a_{1} + a_{90}}{2} \cdot 90 =\]

\[= (10 + 99) \cdot 45 = 4905.\]

\[\textbf{б)}\ a_{1} = 100,\ \ d = 1,\]

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

\[999 = 100 + n - 1\]

\[999 = 99 + n\]

\[n = 900\]

\[S_{900} = \frac{a_{1} + a_{900}}{2} \cdot 900 =\]

\[= (100 + 999) \cdot 450 = 494\ 550.\]