Страница 142 номер 6.2.1, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
По определению арифметической прогрессии \(a_{n+1} = a_n + d\).
1) \(a_1 = 12\), \(d = 3\): \(a_2 = 12 + 3 = 15\), \(a_3 = 15 + 3 = 18\), \(a_4 = 18 + 3 = 21\).
2) \(a_1 = 23\), \(d = 4\): \(a_2 = 23 + 4 = 27\), \(a_3 = 27 + 4 = 31\), \(a_4 = 31 + 4 = 35\).
3) \(a_1 = -98\), \(d = 6{,}5\): \(a_2 = -98 + 6{,}5 = -91{,}5\), \(a_3 = -91{,}5 + 6{,}5 = -85\), \(a_4 = -85 + 6{,}5 = -78{,}5\).
4) \(a_1 = -74\), \(d = 8{,}5\): \(a_2 = -74 + 8{,}5 = -65{,}5\), \(a_3 = -65{,}5 + 8{,}5 = -57\), \(a_4 = -57 + 8{,}5 = -48{,}5\).
5) \(a_1 = 3{,}2\), \(d = -0{,}2\): \(a_2 = 3{,}2 - 0{,}2 = 3\), \(a_3 = 3 - 0{,}2 = 2{,}8\), \(a_4 = 2{,}8 - 0{,}2 = 2{,}6\).
6) \(a_1 = 6{,}3\), \(d = -0{,}3\): \(a_2 = 6{,}3 - 0{,}3 = 6\), \(a_3 = 6 - 0{,}3 = 5{,}7\), \(a_4 = 5{,}7 - 0{,}3 = 5{,}4\).
7) \(a_1 = 1 + \sqrt{3}\), \(d = -\sqrt{3}\): \(a_2 = 1 + \sqrt{3} - \sqrt{3} = 1\), \(a_3 = 1 - \sqrt{3}\), \(a_4 = 1 - \sqrt{3} - \sqrt{3} = 1 - 2\sqrt{3}\).
8) \(a_1 = 1 - \sqrt{2}\), \(d = \sqrt{2}\): \(a_2 = 1 - \sqrt{2} + \sqrt{2} = 1\), \(a_3 = 1 + \sqrt{2}\), \(a_4 = 1 + \sqrt{2} + \sqrt{2} = 1 + 2\sqrt{2}\).
Ответ: 1) \(12;\ 15;\ 18;\ 21\); 2) \(23;\ 27;\ 31;\ 35\); 3) \(-98;\ -91{,}5;\ -85;\ -78{,}5\); 4) \(-74;\ -65{,}5;\ -57;\ -48{,}5\); 5) \(3{,}2;\ 3;\ 2{,}8;\ 2{,}6\); 6) \(6{,}3;\ 6;\ 5{,}7;\ 5{,}4\); 7) \(1 + \sqrt{3};\ 1;\ 1 - \sqrt{3};\ 1 - 2\sqrt{3}\); 8) \(1 - \sqrt{2};\ 1;\ 1 + \sqrt{2};\ 1 + 2\sqrt{2}\).
По определению арифметической прогрессии каждый её член, начиная со второго, получается прибавлением к предыдущему одного и того же числа \(d\): \(a_{n+1} = a_n + d\).
1) \(a_1 = 12\), \(d = 3\): \(a_2 = 12 + 3 = 15\), \(a_3 = 15 + 3 = 18\), \(a_4 = 18 + 3 = 21\).
2) \(a_1 = 23\), \(d = 4\): \(a_2 = 23 + 4 = 27\), \(a_3 = 27 + 4 = 31\), \(a_4 = 31 + 4 = 35\).
3) \(a_1 = -98\), \(d = 6{,}5\): \(a_2 = -98 + 6{,}5 = -91{,}5\), \(a_3 = -91{,}5 + 6{,}5 = -85\), \(a_4 = -85 + 6{,}5 = -78{,}5\).
4) \(a_1 = -74\), \(d = 8{,}5\): \(a_2 = -74 + 8{,}5 = -65{,}5\), \(a_3 = -65{,}5 + 8{,}5 = -57\), \(a_4 = -57 + 8{,}5 = -48{,}5\).
5) \(a_1 = 3{,}2\), \(d = -0{,}2\): \(a_2 = 3{,}2 - 0{,}2 = 3\), \(a_3 = 3 - 0{,}2 = 2{,}8\), \(a_4 = 2{,}8 - 0{,}2 = 2{,}6\).
6) \(a_1 = 6{,}3\), \(d = -0{,}3\): \(a_2 = 6{,}3 - 0{,}3 = 6\), \(a_3 = 6 - 0{,}3 = 5{,}7\), \(a_4 = 5{,}7 - 0{,}3 = 5{,}4\).
7) \(a_1 = 1 + \sqrt{3}\), \(d = -\sqrt{3}\): \(a_2 = 1 + \sqrt{3} - \sqrt{3} = 1\), \(a_3 = 1 - \sqrt{3}\), \(a_4 = 1 - \sqrt{3} - \sqrt{3} = 1 - 2\sqrt{3}\).
8) \(a_1 = 1 - \sqrt{2}\), \(d = \sqrt{2}\): \(a_2 = 1 - \sqrt{2} + \sqrt{2} = 1\), \(a_3 = 1 + \sqrt{2}\), \(a_4 = 1 + \sqrt{2} + \sqrt{2} = 1 + 2\sqrt{2}\).
Ответ: 1) \(12;\ 15;\ 18;\ 21\); 2) \(23;\ 27;\ 31;\ 35\); 3) \(-98;\ -91{,}5;\ -85;\ -78{,}5\); 4) \(-74;\ -65{,}5;\ -57;\ -48{,}5\); 5) \(3{,}2;\ 3;\ 2{,}8;\ 2{,}6\); 6) \(6{,}3;\ 6;\ 5{,}7;\ 5{,}4\); 7) \(1 + \sqrt{3};\ 1;\ 1 - \sqrt{3};\ 1 - 2\sqrt{3}\); 8) \(1 - \sqrt{2};\ 1;\ 1 + \sqrt{2};\ 1 + 2\sqrt{2}\).
