Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 178

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\[1)\ a_{11} = 13;\ \ a_{13} = 7:\]

\[a_{12} = \frac{a_{11} + a_{13}}{2} = \frac{13 + 7}{2} = 10.\]

\[d = a_{13} - a_{12} = 7 - 10 = - 3.\]

\[a_{11} = a_{1} + 10d\]

\[a_{1} = a_{11} - 10d\]

\[a_{1} = 13 - 10 \cdot ( - 3) = 43.\]

\[2)\ a_{11} = - 14;\ \ a_{13} = - 6:\]

\[a_{12} = \frac{a_{11} + a_{13}}{2} =\]

\[= \frac{- 14 - 6}{2} = - 10.\]

\[d = a_{13} - a_{12} = - 6 + 10 = 4.\]

\[a_{11} = a_{1} + 10d\]

\[a_{1} = a_{11} - 10d\]

\[a_{1} = - 14 - 10 \cdot 4 = - 54.\]