Страница 143 номер 6.2.6, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Из формулы \(n\)-го члена \(a_n = a_1 + d(n - 1)\) первый член выражается так: \[a_1 = a_n - d(n - 1).\]
1) \[a_1 = 51{,}8 - (-4) \cdot 12 = 51{,}8 + 48 = 99{,}8.\]
2) \[a_1 = 60{,}3 - (-3) \cdot 15 = 60{,}3 + 45 = 105{,}3.\]
3) \[a_1 = 4 - 0{,}6 \cdot 20 = 4 - 12 = -8.\]
4) \[a_1 = 5 - 0{,}4 \cdot 26 = 5 - 10{,}4 = -5{,}4.\]
5) \[a_1 = -16{,}8 - (-1{,}3) \cdot 30 = -16{,}8 + 39 = 22{,}2.\]
6) \[a_1 = -14{,}3 - (-2{,}1) \cdot 28 = -14{,}3 + 58{,}8 = 44{,}5.\]
Ответ: 1) \(a_1 = 99{,}8\); 2) \(a_1 = 105{,}3\); 3) \(a_1 = -8\); 4) \(a_1 = -5{,}4\); 5) \(a_1 = 22{,}2\); 6) \(a_1 = 44{,}5\).
Из формулы \(n\)-го члена \(a_n = a_1 + d(n - 1)\) первый член выражается так: \[a_1 = a_n - d(n - 1).\]
1) \(d = -4\), \(a_{13} = 51{,}8\): \[a_1 = 51{,}8 - (-4) \cdot 12 = 51{,}8 + 48 = 99{,}8.\]
2) \(d = -3\), \(a_{16} = 60{,}3\): \[a_1 = 60{,}3 - (-3) \cdot 15 = 60{,}3 + 45 = 105{,}3.\]
3) \(d = 0{,}6\), \(a_{21} = 4\): \[a_1 = 4 - 0{,}6 \cdot 20 = 4 - 12 = -8.\]
4) \(d = 0{,}4\), \(a_{27} = 5\): \[a_1 = 5 - 0{,}4 \cdot 26 = 5 - 10{,}4 = -5{,}4.\]
5) \(d = -1{,}3\), \(a_{31} = -16{,}8\): \[a_1 = -16{,}8 - (-1{,}3) \cdot 30 = -16{,}8 + 39 = 22{,}2.\]
6) \(d = -2{,}1\), \(a_{29} = -14{,}3\): \[a_1 = -14{,}3 - (-2{,}1) \cdot 28 = -14{,}3 + 58{,}8 = 44{,}5.\]
Ответ: 1) \(a_1 = 99{,}8\); 2) \(a_1 = 105{,}3\); 3) \(a_1 = -8\); 4) \(a_1 = -5{,}4\); 5) \(a_1 = 22{,}2\); 6) \(a_1 = 44{,}5\).
