Страница 216 номер 738, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
Найдите первый член и разность арифметической прогрессии \((a_n)\), если: 1) \(a_5 + a_{12} = 41\) и \(a_{10} + a_{14} = 62\); 2) \(a_7 + a_{13} = -104\) и \(a_2 \cdot a_6 = -240\).
1) \(a_{5} + a_{12} = 41\) и \(a_{10} + a_{14} = 62\)
\(\begin{cases} a_{1} + 4d + a_{1} + 11d = 41 \\ a_{1} + 9d + a_{1} + 13d = 62 \end{cases}\ \Rightarrow \begin{cases} 2a_{1} + 15d = 41 \\ 2a_{1} + 22d = 62 \end{cases}\ \Rightarrow\)
\(\Rightarrow 7d = 21 \Rightarrow d = 3\)
\(a_{1} = \dfrac{41 - 15d}{2} = \dfrac{41 - 15 \cdot 3}{2} = - \dfrac{4}{2} = - 2\)
2) \(a_{7} + a_{13} = - 104\) и \(a_{2} \cdot a_{6} = - 240\)
\(\begin{cases} a_{1} + 6d + a_{1} + 12d = - 104 \\ \left( a_{1} + d \right)\left( a_{1} + 5d \right) = - 240 \end{cases}\ \Rightarrow \begin{cases} 2a_{1} + 18d = - 104 \\ \left( a_{1} + d \right)\left( a_{1} + 5d \right) = - 240 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} a_{1} = - 52 - 9d \\ ( - 52 - 9d + d)( - 52 - 9d + 5d) = - 240 \end{cases}\)
\(( - 52 - 9d + d)( - 52 - 9d + 5d) = - 240\)
\(( - 52 - 8d)( - 52 - 4d) = - 240\)
\(- 4 \cdot (13 + 2d) \cdot ( - 4)(13 + d) = - 240\)
\((13 + 2d)(13 + d) = - 15\)
\(169 + 26d + 13d + 2d^{2} + 15 = 0\)
\(2d^{2} + 39d + 184 = 0\)
\(D = 1521 - 4 \cdot 2 \cdot 184 = 1521 - 1472 = 49\)
\(d_{1} = \dfrac{- 39 - 7}{4} = - 11{,}5;\ a_{1} = - 52 - 9 \cdot ( - 11{,}5) = 51{,}5\)
\(d_{2} = \dfrac{- 39 + 7}{4} = - 8;\ a_{1} = - 52 - 9 \cdot ( - 8) = 20\)
Ответ: 1) \(d = 3;\ a_{1} = - 2\); 2) \(d_{1} = - 11{,}5;\ a_{1} = 51{,}5\); \(d_{2} = - 8;\ a_{1} = 20\).
