Страницы 142-143 номер 6.2.3, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Член с данным номером находим по формуле \(n\)-го члена арифметической прогрессии \[a_n = a_1 + d(n - 1).\]
1) \[a_{10} = 7 + 5 \cdot (10 - 1) = 7 + 45 = 52.\]
2) \[a_{12} = 9 + 4 \cdot (12 - 1) = 9 + 44 = 53.\]
3) \[a_{19} = -20 + \frac{1}{2} \cdot (19 - 1) = -20 + 9 = -11.\]
4) \[a_{22} = -10 + \frac{1}{3} \cdot (22 - 1) = -10 + 7 = -3.\]
5) \[a_{31} = 0{,}5 + (-0{,}1) \cdot (31 - 1) = 0{,}5 - 3 = -2{,}5.\]
6) \[a_{29} = 0{,}4 + (-0{,}2) \cdot (29 - 1) = 0{,}4 - 5{,}6 = -5{,}2.\]
7) \[a_{16} = -53 + (-2) \cdot (16 - 1) = -53 - 30 = -83.\]
8) \[a_{25} = -102 + (-1) \cdot (25 - 1) = -102 - 24 = -126.\]
Ответ: 1) \(a_{10} = 52\); 2) \(a_{12} = 53\); 3) \(a_{19} = -11\); 4) \(a_{22} = -3\); 5) \(a_{31} = -2{,}5\); 6) \(a_{29} = -5{,}2\); 7) \(a_{16} = -83\); 8) \(a_{25} = -126\).
Член прогрессии с данным номером находится по формуле \(n\)-го члена \[a_n = a_1 + d(n - 1).\]
1) \(a_1 = 7\), \(d = 5\), найти \(a_{10}\): \[a_{10} = 7 + 5 \cdot (10 - 1) = 7 + 45 = 52.\]
2) \(a_1 = 9\), \(d = 4\), найти \(a_{12}\): \[a_{12} = 9 + 4 \cdot (12 - 1) = 9 + 44 = 53.\]
3) \(a_1 = -20\), \(d = \frac{1}{2}\), найти \(a_{19}\): \[a_{19} = -20 + \frac{1}{2} \cdot (19 - 1) = -20 + 9 = -11.\]
4) \(a_1 = -10\), \(d = \frac{1}{3}\), найти \(a_{22}\): \[a_{22} = -10 + \frac{1}{3} \cdot (22 - 1) = -10 + 7 = -3.\]
5) \(a_1 = 0{,}5\), \(d = -0{,}1\), найти \(a_{31}\): \[a_{31} = 0{,}5 + (-0{,}1) \cdot (31 - 1) = 0{,}5 - 3 = -2{,}5.\]
6) \(a_1 = 0{,}4\), \(d = -0{,}2\), найти \(a_{29}\): \[a_{29} = 0{,}4 + (-0{,}2) \cdot (29 - 1) = 0{,}4 - 5{,}6 = -5{,}2.\]
7) \(a_1 = -53\), \(d = -2\), найти \(a_{16}\): \[a_{16} = -53 + (-2) \cdot (16 - 1) = -53 - 30 = -83.\]
8) \(a_1 = -102\), \(d = -1\), найти \(a_{25}\): \[a_{25} = -102 + (-1) \cdot (25 - 1) = -102 - 24 = -126.\]
Ответ: 1) \(a_{10} = 52\); 2) \(a_{12} = 53\); 3) \(a_{19} = -11\); 4) \(a_{22} = -3\); 5) \(a_{31} = -2{,}5\); 6) \(a_{29} = -5{,}2\); 7) \(a_{16} = -83\); 8) \(a_{25} = -126\).
