Дана арифметическая прогрессия, где a_1+a_3+a_5=-12; a_1*a_3*a_5=80. Найдите a_1.

Ответ:

\[a_{1} + a_{3} + a_{5} = - 12;\]

\[a_{1} \cdot a_{3} \cdot a_{5} = 80\]

\[a_{1} + a_{1} + 2d + a_{1} + 4d = - 12\]

\[3a_{1} + 6d = - 12\]

\[a_{1} + 2d = - 4;\]

\[a_{1} = - 4 - 2d.\]

\[a_{1}\left( a_{1} + 2d \right)\left( a_{1} + 4d \right) = 80\]

\[( - 4 - 2d)( - 4 - 2d + 2d)( - 4 - 2d + 4d) = 80\]

\[( - 4 - 2d)( - 4)( - 4 + 2d) = 80\]

\[- (4 + 2d)(2d - 4) = - 20\]

\[4d^{2} - 16 = 20\]

\[4d^{2} = 36\]

\[d^{2} = 9\]

\[d = \pm 3.\]

\[a_{1} = - 4 - 2 \cdot 3 = - 10;\]

\[a_{1} = - 4 - 2 \cdot ( - 3) = 2.\]

\[Ответ:a_{1} = - 10\ или\ a_{1} = 2.\]