ГДЗ по Алгебре 9 Класс Номер 1031 Мерзляк, Полонский, Якир — Подробные Ответы

Пусть \(S_n\) — сумма \(n\) первых членов арифметической прогрессии \((a_n)\). Найдите первый член и разность прогрессии, если:

1) \(a_1 + a_2 + a_3 = 18\), \(a_4 + a_5 = -2;\)

2) \(a_1 — a_3 = -4\), \(a_1 a_4 = -3;\)

3) \(a_2 + a_4 + a_6 = 36\), \(a_1 a_5 = 54;\)

4) \(S_5 — S_2 — a_1 = 0,1\), \(a_1 + S_4 = 0,1;\)

5) \(S_5 = 9\), \(S_8 = 22,5.\)

\(a_1 = -7,\, d = 3\)

\(a_1 = 3,\, a_1 = 5,\, d = -2\)

\(a_1 = 3,\, d = 3\) и \(a_1 = -33,\, d = 15\)

\(a_1 = -0{,}7,\, d = 0{,}3\)

\(a_1 = 0,\, d = 1{,}5\)

1. Пусть \(a_1\) — первый член, \(d\) — разность. Запишем:
\(a_1 + a_2 + a_3 = 18\), \(a_4 + a_5 = -2\).
\(a_2 = a_1 + d\), \(a_3 = a_1 + 2d\), \(a_4 = a_1 + 3d\), \(a_5 = a_1 + 4d\).
Составим уравнения: \(a_1 + (a_1 + d) + (a_1 + 2d) = 18\), то есть \(3a_1 + 3d = 18\), откуда \(a_1 + d = 6\).
Второе: \((a_1 + 3d) + (a_1 + 4d) = -2\), то есть \(2a_1 + 7d = -2\).
Подставим \(a_1 = 6 — d\) во второе: \(2(6 — d) + 7d = -2\), \(12 — 2d + 7d = -2\), \(12 + 5d = -2\), \(5d = -14\), \(d = -\frac{14}{5} = -2{,}8\).
\(a_1 = 6 — (-2{,}8) = 8{,}8\).

2. \(a_1 — a_3 = -4\), \(a_1 a_4 = -3\).
\(a_3 = a_1 + 2d\), \(a_4 = a_1 + 3d\).
\(a_1 — (a_1 + 2d) = -4\), \(-2d = -4\), \(d = 2\).
\(a_1(a_1 + 3d) = -3\), \(a_1(a_1 + 6) = -3\), \(a_1^2 + 6a_1 + 3 = 0\).
Решаем: \(a_1 = \frac{-6 \pm \sqrt{36 — 12}}{2} = \frac{-6 \pm \sqrt{24}}{2} = \frac{-6 \pm 2\sqrt{6}}{2} = -3 \pm \sqrt{6}\).

3. \(a_2 + a_4 + a_6 = 36\), \(a_1 a_5 = 54\).
\(a_2 = a_1 + d\), \(a_4 = a_1 + 3d\), \(a_6 = a_1 + 5d\), \(a_5 = a_1 + 4d\).
Сумма: \(a_2 + a_4 + a_6 = (a_1 + d) + (a_1 + 3d) + (a_1 + 5d) = 3a_1 + 9d = 36\), \(a_1 + 3d = 12\).
\(a_1 a_5 = a_1(a_1 + 4d) = 54\).
\(a_1 = 12 — 3d\), подставим: \((12 — 3d)(12 + d) = 54\), \(144 + 12d — 36d — 3d^2 = 54\), \(-3d^2 — 24d + 90 = 0\), \(3d^2 + 24d — 90 = 0\), \(d^2 + 8d — 30 = 0\).
\(d = \frac{-8 \pm \sqrt{64 + 120}}{2} = \frac{-8 \pm \sqrt{184}}{2}\), \(\sqrt{184} \approx 13{,}56\).
\(d_1 = \frac{-8 + 13{,}56}{2} \approx 2{,}78\), \(d_2 = \frac{-8 — 13{,}56}{2} \approx -10{,}78\).
\(a_1 = 12 — 3d\): для \(d_1\) \(a_1 \approx 12 — 8{,}34 = 3{,}66\), для \(d_2\) \(a_1 \approx 12 + 32{,}34 = 44{,}34\).

4. \(S_5 — S_2 — a_1 = 0{,}1\), \(a_1 + S_4 = 0{,}1\).
\(S_n = \frac{2a_1 + (n-1)d}{2} \cdot n\).
\(S_5 = (a_1 + 2d) \cdot 5\), \(S_2 = (a_1 + 0{,}5d) \cdot 2\).
\(5(a_1 + 2d) — 2(a_1 + 0{,}5d) — a_1 = 0{,}1\), \(5a_1 + 10d — 2a_1 — d — a_1 = 0{,}1\), \(2a_1 + 9d = 0{,}1\).
\(S_4 = (a_1 + 1{,}5d) \cdot 4\), \(a_1 + 4(a_1 + 1{,}5d) = 0{,}1\), \(a_1 + 4a_1 + 6d = 0{,}1\), \(5a_1 + 6d = 0{,}1\).
Система: \(2a_1 + 9d = 0{,}1\), \(5a_1 + 6d = 0{,}1\).
Умножим первое на 5: \(10a_1 + 45d = 0{,}5\), второе на 2: \(10a_1 + 12d = 0{,}2\).
Вычтем: \(33d = 0{,}3\), \(d = \frac{0{,}3}{33} \approx 0{,}0091\).
\(5a_1 + 6 \cdot 0{,}0091 = 0{,}1\), \(5a_1 = 0{,}1 — 0{,}0546 = 0{,}0454\), \(a_1 \approx 0{,}0091\).

5. \(S_5 = 9\), \(S_8 = 22{,}5\).
\(S_5 = (a_1 + 2d) \cdot 5 = 9\), \(a_1 + 2d = \frac{9}{5} = 1{,}8\).
\(S_8 = (a_1 + 3{,}5d) \cdot 8 = 22{,}5\), \(a_1 + 3{,}5d = \frac{22{,}5}{8} = 2{,}8125\).
\(a_1 + 3{,}5d — (a_1 + 2d) = 2{,}8125 — 1{,}8\), \(1{,}5d = 1{,}0125\), \(d = \frac{1{,}0125}{1{,}5} = 0{,}675\).
\(a_1 = 1{,}8 — 2 \cdot 0{,}675 = 1{,}8 — 1{,}35 = 0{,}45\).