Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1451

\[\boxed{\mathbf{1451}\mathbf{.}}\]

\[a_{5} \bullet a_{6} = 33\left( a_{1} \bullet a_{2} \right)\]

\[\left( a_{1} + 4d \right)\left( a_{1} + 5d \right) =\]

\[= 33a_{1}\left( a_{1} + d \right)\]

\[a_{1}^{2} + 5da_{1} + 4da_{1} + 20d^{2} =\]

\[= 33a_{1}^{2} + 33da_{1}\]

\[32a_{1}^{2} + 24a_{1} - 20d^{2} = 0\]

\[8a_{1}^{2} + 6da_{1} - 5d^{2} = 0\]

\[D = (6d)^{2} + 4 \bullet 8 \bullet 5d^{2} =\]

\[= 36d^{2} + 160d^{2} = 196d^{2}\]

\[a_{11} = \frac{- 6d - \sqrt{196d^{2}}}{2 \bullet 8} =\]

\[= \frac{- 6d - 14d}{16} = \frac{- 20d}{16} = - \frac{5d}{4};\]

\[a_{12} = \frac{- 6d + \sqrt{196d^{2}}}{2 \bullet 8} =\]

\[= \frac{- 6d + 14d}{16} = \frac{8d}{16} = \frac{d}{2}.\]

\[Все\ члены\ арифметической\ \]

\[прогрессии\ положительны:\]

\[a_{1} = \frac{d}{2}.\]

\[Найдем\ отношение:\]

\[\frac{a_{5}}{a_{2}} = \frac{a_{1} + 4d}{a_{1} + d} = \frac{\frac{d}{2} + 4d}{\frac{d}{2} + d} =\]

\[= \frac{9d}{2}\ :\frac{3d}{2} = \frac{9d}{3d} = 3.\]

\[Ответ:\ \ в\ 3\ раза.\]