Предмет: Алгебра, автор: Аноним

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Приложения:

Ответы

Автор ответа: FaerVator
2

Ответ:

b₇ + b₈ + b₉ = 18954

Объяснение:

Дано:

b_n - геом.прогрессия

n = 9

b₁ + b₂ + b₃ = 26

b₄ + b₅ + b₆ = 702

Найти: b₇ + b₈ + b₉

Решение:

Составим систему уравнений:

\displaystyle \left \{ {{b_1+b_2+b_3=26} \atop {b_4+b_5+b_6=702}} \right.

Каждый член представим по формуле bₙ = b₁qⁿ⁻¹:

\displaystyle \left \{ {{b_1+b_1q+b_1q^2=26} \atop {b_1q^3+b_1q^4+b_1q^5=702}} \right. \Rightarrow\left \{ {{b_1+b_1q+b_1q^2=26} \atop {q^3(b_1+b_1q+b_1q^2)=702}} \right.

Разделим второе уравнение на первое:

q^3=27\\q=\sqrt[3]{27} =3

Из формулы суммы n-первых членов геом.прогрессии S_n=\frac{b_1(q^n-1)}{q-1}  найдём b₁.

Пусть в нашем случае n = 3 , т.к сумма первых трёх членов геом.прогрессии нам известна:

\displaystyle S_3=\frac{b_1(q^n-1)}{q-1} \\26=\frac{b_1(3^3-1)}{3-1}\\26=\frac{26b_1}{2} \\26b_1=26\cdot 2\\26b_1=52\\b_1=52:26 =2

Найдём сумму седьмого , восьмого и девятого члена геом.прогрессии:

b₇ + b₈ + b₉ =b₁q⁶ + b₁q⁷ + b₁q⁸ = 2 · 3⁶ + 2 · 3⁷ + 2 · 3⁸ = 1458 + 4374 + 13122 = 18954

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