Страница 218 номер 752, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

\(a_{n} = \dfrac{a_{n - 1} + a_{n + 1}}{2} \Rightarrow a_{2} = \dfrac{a_{1} + a_{3}}{2};\ a_{3} = \dfrac{a_{2} + a_{4}}{2}\)
\(3y + 5 = \dfrac{y^{2} - 2y + 4y + 13}{2}\)
\(6y + 10 = y^{2} + 2y + 13\)
\(y^{2} - 4y + 3 = 0\)
\((y - 1)(y - 3) = 0\)
\(y_{1} = 1;\ y_{2} = 3\)
\(4y + 13 = \dfrac{3y + 5 + 2y^{2} - y + 25}{2}\)
\(8y + 26 = 2y^{2} + 2y + 30\)
\(2y^{2} - 6y + 4 = 0\)
\(y^{2} - 3y + 2 = 0\)
\((y - 1)(y - 2) = 0\)
\(y_{1} = 1;\ y_{2} = 2\)
\(\begin{cases} y_{1} = 1;\ y_{2} = 3 \\ y_{1} = 1;\ y_{2} = 2 \end{cases}\ \Rightarrow y = 1\)
\(a_{1} = y^{2} - 2y = 1^{2} - 2 \cdot 1 = 1 - 2 = - 1\)
\(a_{2} = 3 \cdot y + 5 = 3 \cdot 1 + 5 = 8\)
\(a_{3} = 4y + 13 = 4 \cdot 1 + 13 = 17\)
\(a_{4} = 2y^{2} - y + 25 = 2 - 1 + 25 = 26\)

Ответ: \(y = 1\); \(a_{1} = - 1\), \(a_{2} = 8\), \(a_{3} = 17\), \(a_{4} = 26\).