Решебник по алгебре 9 класс Мерзляк Задание 705

\[\boxed{\mathbf{705\ (705).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[b_{n} = - n^{2} + 15n - 20\]

\[- n^{2} + 15n - 20 > 16\ \ \]

\[- n^{2} + 15n - 36 > 0\]

\[n^{2} - 15n + 36 < 0\]

\[n_{1} + n_{2} = 15,\ \ n_{1} = 12\]

\[n_{1}n_{2} = 36,\ \ n_{2} = 3\]

\[Тогда\ 4;5;6;7;8;9;10;\]

\[11 - 8\ членов.\]

\[Ответ:8.\]

\[\boxed{\mathbf{705.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 12,\ 72,\ 432,\ \ldots,\ \ \]

\[b_{1} = 12;\ \ q = \frac{72}{12} = 6:\]

\[S_{5} = \frac{12 \cdot \left( 6^{5} - 1 \right)}{6 - 1} =\]

\[= \frac{12 \cdot 7775}{5} = 12 \cdot 1555 =\]

\[= 18\ 660.\]

\[2)\ \frac{1}{16};\ - \frac{1}{8};\ \frac{1}{4};\ldots\text{\ \ }\]

\[b_{1} = \frac{1}{16};\ \ \ q = \frac{- 1 \cdot 16}{8 \cdot 1} = - 2:\]

\[S_{5} = \frac{\frac{1}{16} \cdot \left( ( - 2)^{5} - 1 \right)}{- 2 - 1} =\]

\[= \frac{\frac{1}{16} \cdot ( - 33)}{- 3} = \frac{1}{16} \cdot 11 = \frac{11}{16}.\]