Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 982

\[\left( a_{n} \right) - арифметическая\ \]

\[прогрессия:\]

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

\[\left( a_{1} + 4d \right)\left( a_{1} + 5d \right) = 33a_{1}\left( a_{1} + d \right)\]

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

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

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

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

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

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

\[a_{1,1} = \frac{- 6d - 14d}{16} = - \frac{5d}{4};\]

\[a_{1,2} = \frac{- 6d + 14d}{16} = \frac{d}{2}.\]

\[1)\ Члены\ положительны:\]

\[a_{5} > a_{1};\ \ \ d > 0;\ \ \ a_{1} = \frac{d}{2}.\]

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

\[= \frac{d + 8d}{d + 2d} = \frac{9d}{3d} = 3.\]

\[Ответ:\ \ в\ три\ раза.\]