Страница 142 номер 6.2.2, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Разность прогрессии \(d = a_2 - a_1\), а каждый следующий член получается прибавлением \(d\) к предыдущему.
1) \(d = 9 - 5 = 4\); \(a_3 = 9 + 4 = 13\), \(a_4 = 13 + 4 = 17\), \(a_5 = 17 + 4 = 21\).
2) \(d = 19 - 13 = 6\); \(a_3 = 19 + 6 = 25\), \(a_4 = 25 + 6 = 31\), \(a_5 = 31 + 6 = 37\).
3) \(d = -6 - (-8) = 2\); \(a_3 = -6 + 2 = -4\), \(a_4 = -4 + 2 = -2\), \(a_5 = -2 + 2 = 0\).
4) \(d = -9 - (-12) = 3\); \(a_3 = -9 + 3 = -6\), \(a_4 = -6 + 3 = -3\), \(a_5 = -3 + 3 = 0\).
5) \(d = -2 - 0{,}5 = -2{,}5\); \(a_3 = -2 - 2{,}5 = -4{,}5\), \(a_4 = -4{,}5 - 2{,}5 = -7\), \(a_5 = -7 - 2{,}5 = -9{,}5\).
6) \(d = -1{,}5 - 0{,}3 = -1{,}8\); \(a_3 = -1{,}5 - 1{,}8 = -3{,}3\), \(a_4 = -3{,}3 - 1{,}8 = -5{,}1\), \(a_5 = -5{,}1 - 1{,}8 = -6{,}9\).
7) \(d = \frac{1}{3} - \frac{5}{6} = \frac{2}{6} - \frac{5}{6} = -\frac{1}{2}\); \(a_3 = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}\), \(a_4 = -\frac{1}{6} - \frac{1}{2} = -\frac{2}{3}\), \(a_5 = -\frac{2}{3} - \frac{1}{2} = -1\frac{1}{6}\).
8) \(d = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}\); \(a_3 = \frac{3}{4} + \frac{1}{4} = 1\), \(a_4 = 1 + \frac{1}{4} = 1\frac{1}{4}\), \(a_5 = 1\frac{1}{4} + \frac{1}{4} = 1\frac{1}{2}\).
Ответ: 1) \(a_3 = 13\), \(a_4 = 17\), \(a_5 = 21\); 2) \(a_3 = 25\), \(a_4 = 31\), \(a_5 = 37\); 3) \(a_3 = -4\), \(a_4 = -2\), \(a_5 = 0\); 4) \(a_3 = -6\), \(a_4 = -3\), \(a_5 = 0\); 5) \(a_3 = -4{,}5\), \(a_4 = -7\), \(a_5 = -9{,}5\); 6) \(a_3 = -3{,}3\), \(a_4 = -5{,}1\), \(a_5 = -6{,}9\); 7) \(a_3 = -\frac{1}{6}\), \(a_4 = -\frac{2}{3}\), \(a_5 = -1\frac{1}{6}\); 8) \(a_3 = 1\), \(a_4 = 1\frac{1}{4}\), \(a_5 = 1\frac{1}{2}\).
Разность арифметической прогрессии равна разности между любым её членом, начиная со второго, и предыдущим членом, поэтому \(d = a_2 - a_1\). Дальше каждый следующий член получается прибавлением \(d\) к предыдущему.
1) \(d = 9 - 5 = 4\); \(a_3 = 9 + 4 = 13\), \(a_4 = 13 + 4 = 17\), \(a_5 = 17 + 4 = 21\).
2) \(d = 19 - 13 = 6\); \(a_3 = 19 + 6 = 25\), \(a_4 = 25 + 6 = 31\), \(a_5 = 31 + 6 = 37\).
3) \(d = -6 - (-8) = 2\); \(a_3 = -6 + 2 = -4\), \(a_4 = -4 + 2 = -2\), \(a_5 = -2 + 2 = 0\).
4) \(d = -9 - (-12) = 3\); \(a_3 = -9 + 3 = -6\), \(a_4 = -6 + 3 = -3\), \(a_5 = -3 + 3 = 0\).
5) \(d = -2 - 0{,}5 = -2{,}5\); \(a_3 = -2 - 2{,}5 = -4{,}5\), \(a_4 = -4{,}5 - 2{,}5 = -7\), \(a_5 = -7 - 2{,}5 = -9{,}5\).
6) \(d = -1{,}5 - 0{,}3 = -1{,}8\); \(a_3 = -1{,}5 - 1{,}8 = -3{,}3\), \(a_4 = -3{,}3 - 1{,}8 = -5{,}1\), \(a_5 = -5{,}1 - 1{,}8 = -6{,}9\).
7) \(d = \frac{1}{3} - \frac{5}{6} = \frac{2}{6} - \frac{5}{6} = -\frac{1}{2}\); \(a_3 = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}\), \(a_4 = -\frac{1}{6} - \frac{1}{2} = -\frac{2}{3}\), \(a_5 = -\frac{2}{3} - \frac{1}{2} = -1\frac{1}{6}\).
8) \(d = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}\); \(a_3 = \frac{3}{4} + \frac{1}{4} = 1\), \(a_4 = 1 + \frac{1}{4} = 1\frac{1}{4}\), \(a_5 = 1\frac{1}{4} + \frac{1}{4} = 1\frac{1}{2}\).
Ответ: 1) \(a_3 = 13\), \(a_4 = 17\), \(a_5 = 21\); 2) \(a_3 = 25\), \(a_4 = 31\), \(a_5 = 37\); 3) \(a_3 = -4\), \(a_4 = -2\), \(a_5 = 0\); 4) \(a_3 = -6\), \(a_4 = -3\), \(a_5 = 0\); 5) \(a_3 = -4{,}5\), \(a_4 = -7\), \(a_5 = -9{,}5\); 6) \(a_3 = -3{,}3\), \(a_4 = -5{,}1\), \(a_5 = -6{,}9\); 7) \(a_3 = -\frac{1}{6}\), \(a_4 = -\frac{2}{3}\), \(a_5 = -1\frac{1}{6}\); 8) \(a_3 = 1\), \(a_4 = 1\frac{1}{4}\), \(a_5 = 1\frac{1}{2}\).
