Предмет: Алгебра, автор: novichek21

Задание во вложении.

Приложения:

Ответы

Автор ответа: сок111213

2

$b_{3} + b_{4} = 4 .5\\ b_{5} - b_{3} = - 2.25 \\ \\ b_{n} = b_{1} {q}^{n - 1} \\ \\ b_{1} {q}^{2} + b_{1} {q}^{3} = 4.5 \\ b_{1} {q}^{4} - b_{1} {q}^{2} = - 2.25 \\ \\ b_{1} {q}^{2} (1 + q) = 4.5 \\ b_{1} {q}^{2} ( {q}^{2} - 1) = - 2.25 \\ \\ b_{1} {q}^{2} (1 + q) = 4.5 \\ b_{1} {q}^{2} ( {q}- 1)(q + 1) = - 2.25$

Разделим первое уравнений на второе:

$\frac{b_{1} {q}^{2} (1 + q) }{b_{1} {q}^{2} ( {q} - 1)(q + 1) } = \frac{4.5}{ - 2.25} \\ \frac{1}{q - 1} = - \frac{9 \times 4}{2 \times 9} \\ \frac{1}{q - 1} = - 2 \\ - 2(q - 1) = 1 \\ q - 1 = - 0.5 \\ q = - 0.5 + 1 \\ q = 0.5$

Полставим это значение в первое уравнение:

$b_{1} \times { (\frac{1}{2}) }^{2} + b_{1} \times {( \frac{1}{2}) }^{3} = 4.5 \\ \frac{1}{4} b_{1} + \frac{1}{8} b_{1} = 4.5 \\ \frac{2}{8} b_{1} + \frac{1}{8} b_{1} = \frac{9}{2} \\ \frac{3}{8} b_{1} = \frac{9}{2} \\ b_{1} = \frac{9}{2} \div \frac{3}{8} \\ b_{1} = \frac{9 \times 8}{2 \times 3} \\ b_{1} = 12$

Формула суммы n первых членов геометрической прогрессии:

$s_{n} = \frac{b_{1}( {q}^{n} - 1)}{q - 1}$

Для n = 5 :

$s_{5} = \frac{b_{1}( {q}^{5} - 1)}{q - 1} = \frac{12(( \frac{1}{2}) {}^{5} - 1)}{ \frac{1}{2} - 1} = \\ \frac{12( \frac{1}{32} - \frac{32}{32} )}{ - \frac{1}{2} } = - 24 \times ( - \frac{31}{32} ) = \\ \frac{3 \times 31}{4} = \frac{93 }{4} = 23.25$

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