7класс

Страница 148 номер 6.3.1, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой

§6. Прогрессии. 6.3. Геометрическая прогрессия. Страница 148. Номер 6.3.1

Решение:

1) \(b_2 = 5 \cdot 3 = 15\), \(b_3 = 15 \cdot 3 = 45\), \(b_4 = 45 \cdot 3 = 135\), \(b_5 = 135 \cdot 3 = 405\).

2) \(b_2 = 4 \cdot 2 = 8\), \(b_3 = 8 \cdot 2 = 16\), \(b_4 = 16 \cdot 2 = 32\), \(b_5 = 32 \cdot 2 = 64\).

3) \(b_2 = -6 \cdot \frac{1}{2} = -3\), \(b_3 = -3 \cdot \frac{1}{2} = -\frac{3}{2} = -1\frac{1}{2}\), \(b_4 = -\frac{3}{2} \cdot \frac{1}{2} = -\frac{3}{4}\), \(b_5 = -\frac{3}{4} \cdot \frac{1}{2} = -\frac{3}{8}\).

4) \(b_2 = -12 \cdot \frac{1}{3} = -4\), \(b_3 = -4 \cdot \frac{1}{3} = -\frac{4}{3} = -1\frac{1}{3}\), \(b_4 = -\frac{4}{3} \cdot \frac{1}{3} = -\frac{4}{9}\), \(b_5 = -\frac{4}{9} \cdot \frac{1}{3} = -\frac{4}{27}\).

5) \(b_2 = 27 \cdot \left(-\frac{1}{3}\right) = -9\), \(b_3 = -9 \cdot \left(-\frac{1}{3}\right) = 3\), \(b_4 = 3 \cdot \left(-\frac{1}{3}\right) = -1\), \(b_5 = -1 \cdot \left(-\frac{1}{3}\right) = \frac{1}{3}\).

6) \(b_2 = 16 \cdot \left(-\frac{1}{2}\right) = -8\), \(b_3 = -8 \cdot \left(-\frac{1}{2}\right) = 4\), \(b_4 = 4 \cdot \left(-\frac{1}{2}\right) = -2\), \(b_5 = -2 \cdot \left(-\frac{1}{2}\right) = 1\).

7) \(b_2 = -\frac{1}{8} \cdot (-2) = \frac{1}{4}\), \(b_3 = \frac{1}{4} \cdot (-2) = -\frac{1}{2}\), \(b_4 = -\frac{1}{2} \cdot (-2) = 1\), \(b_5 = 1 \cdot (-2) = -2\).

8) \(b_2 = -\frac{1}{9} \cdot (-3) = \frac{1}{3}\), \(b_3 = \frac{1}{3} \cdot (-3) = -1\), \(b_4 = -1 \cdot (-3) = 3\), \(b_5 = 3 \cdot (-3) = -9\).

Ответ: 1) \(5;\ 15;\ 45;\ 135;\ 405\); 2) \(4;\ 8;\ 16;\ 32;\ 64\); 3) \(-6;\ -3;\ -1\frac{1}{2};\ -\frac{3}{4};\ -\frac{3}{8}\); 4) \(-12;\ -4;\ -1\frac{1}{3};\ -\frac{4}{9};\ -\frac{4}{27}\); 5) \(27;\ -9;\ 3;\ -1;\ \frac{1}{3}\); 6) \(16;\ -8;\ 4;\ -2;\ 1\); 7) \(-\frac{1}{8};\ \frac{1}{4};\ -\frac{1}{2};\ 1;\ -2\); 8) \(-\frac{1}{9};\ \frac{1}{3};\ -1;\ 3;\ -9\).

Решение:

По определению геометрической прогрессии каждый её член, начиная со второго, равен предыдущему члену, умноженному на знаменатель \(q\).

1) \(b_1 = 5\), \(q = 3\): \(b_2 = 5 \cdot 3 = 15\), \(b_3 = 15 \cdot 3 = 45\), \(b_4 = 45 \cdot 3 = 135\), \(b_5 = 135 \cdot 3 = 405\).

2) \(b_1 = 4\), \(q = 2\): \(b_2 = 4 \cdot 2 = 8\), \(b_3 = 8 \cdot 2 = 16\), \(b_4 = 16 \cdot 2 = 32\), \(b_5 = 32 \cdot 2 = 64\).

3) \(b_1 = -6\), \(q = \frac{1}{2}\): \(b_2 = -6 \cdot \frac{1}{2} = -3\), \(b_3 = -3 \cdot \frac{1}{2} = -\frac{3}{2} = -1\frac{1}{2}\), \(b_4 = -\frac{3}{2} \cdot \frac{1}{2} = -\frac{3}{4}\), \(b_5 = -\frac{3}{4} \cdot \frac{1}{2} = -\frac{3}{8}\).

4) \(b_1 = -12\), \(q = \frac{1}{3}\): \(b_2 = -12 \cdot \frac{1}{3} = -4\), \(b_3 = -4 \cdot \frac{1}{3} = -\frac{4}{3} = -1\frac{1}{3}\), \(b_4 = -\frac{4}{3} \cdot \frac{1}{3} = -\frac{4}{9}\), \(b_5 = -\frac{4}{9} \cdot \frac{1}{3} = -\frac{4}{27}\).

5) \(b_1 = 27\), \(q = -\frac{1}{3}\): \(b_2 = 27 \cdot \left(-\frac{1}{3}\right) = -9\), \(b_3 = -9 \cdot \left(-\frac{1}{3}\right) = 3\), \(b_4 = 3 \cdot \left(-\frac{1}{3}\right) = -1\), \(b_5 = -1 \cdot \left(-\frac{1}{3}\right) = \frac{1}{3}\).

6) \(b_1 = 16\), \(q = -\frac{1}{2}\): \(b_2 = 16 \cdot \left(-\frac{1}{2}\right) = -8\), \(b_3 = -8 \cdot \left(-\frac{1}{2}\right) = 4\), \(b_4 = 4 \cdot \left(-\frac{1}{2}\right) = -2\), \(b_5 = -2 \cdot \left(-\frac{1}{2}\right) = 1\).

7) \(b_1 = -\frac{1}{8}\), \(q = -2\): \(b_2 = -\frac{1}{8} \cdot (-2) = \frac{1}{4}\), \(b_3 = \frac{1}{4} \cdot (-2) = -\frac{1}{2}\), \(b_4 = -\frac{1}{2} \cdot (-2) = 1\), \(b_5 = 1 \cdot (-2) = -2\).

8) \(b_1 = -\frac{1}{9}\), \(q = -3\): \(b_2 = -\frac{1}{9} \cdot (-3) = \frac{1}{3}\), \(b_3 = \frac{1}{3} \cdot (-3) = -1\), \(b_4 = -1 \cdot (-3) = 3\), \(b_5 = 3 \cdot (-3) = -9\).