Страница 216 номер 737, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
Найдите первый член и разность арифметической прогрессии \((a_n)\), если: 1) \(a_3 + a_7 = 30\) и \(a_6 + a_{16} = 60\); 2) \(a_4 + a_{10} = 36\) и \(a_5 \cdot a_{11} = 340\).
\(a_{n} = a_{1} + d(n - 1)\)
1) \(a_{3} + a_{7} = 30\) и \(a_{6} + a_{16} = 60\)
\(\begin{cases} a_{3} + a_{7} = 30 \\ a_{6} + a_{16} = 60 \end{cases}\ \Rightarrow \begin{cases} a_{1} + 2d + a_{1} + 6d = 30 \\ a_{1} + 5d + a_{1} + 15d = 60 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} 2a_{1} + 8d = 30 \\ 2a_{1} + 20d = 60 \end{cases}\ \Rightarrow 12d = 30 \Rightarrow d = 2{,}5\)
\(a_{1} = \dfrac{30 - 8d}{2} = 15 - 4 \cdot 2{,}5 = 5\)
2) \(a_{4} + a_{10} = 36\) и \(a_{5} \cdot a_{11} = 340\)
\(\begin{cases} a_{4} + a_{10} = 36 \\ a_{5} \cdot a_{11} = 340 \end{cases}\ \Rightarrow \begin{cases} a_{1} + 3d + a_{1} + 9d = 36 \\ \left( a_{1} + 4d \right)\left( a_{1} + 10d \right) = 340 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} 2a_{1} + 12d = 36 \\ \left( a_{1} + 4d \right)\left( a_{1} + 10d \right) = 340 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} a_{1} = 18 - 6d \\ (18 - 6d + 4d)(18 - 6d + 10d) = 340 \end{cases}\ \Rightarrow\)
\((18 - 6d + 4d)(18 - 6d + 10d) = 340\)
\((18 - 2d)(18 + 4d) = 340\)
\(324 - 36d + 72d - 8d^{2} = 340\)
\(8d^{2} - 36d + 16 = 0\)
\(2d^{2} - 9d + 4 = 0\)
\(D = 81 - 4 \cdot 2 \cdot 4 = 81 - 32 = 49\)
\(d_{1} = \dfrac{9 - 7}{4} = \dfrac{2}{4} = \dfrac{1}{2} \Rightarrow a_{1} = 18 - 6 \cdot \dfrac{1}{2} = 18 - 3 = 15\)
\(d_{2} = \dfrac{9 + 7}{4} = 4 \Rightarrow a_{1} = 18 - 6 \cdot 4 = 18 - 24 = - 6\)
Ответ: 1) \(d = 2{,}5;\ a_{1} = 5\); 2) \(d = \frac{1}{2};\ a_{1} = 15;\ d = 4;\ a_{1} = - 6\).
