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greenwalley
@greenwalley
August 2022
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Знаменатель геометрической прогрессии равен 2/3,а сумма четырёх первых членов равна 65.Найдите первый член прогрессии.
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nafanya2014
Verified answer
B₂=b₁q=(2/3)b₁
b₃=b₂q=(4/9)b₁
b₄=b₃q=(8/27)b₁
S₄=b₁+b₂+b₃+b₄
65=b₁·(1+(2/3)+(4/9)+(8/27)
65=b₁·(27+18+12+8)/27
b₁=27
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Answers & Comments
Verified answer
B₂=b₁q=(2/3)b₁b₃=b₂q=(4/9)b₁
b₄=b₃q=(8/27)b₁
S₄=b₁+b₂+b₃+b₄
65=b₁·(1+(2/3)+(4/9)+(8/27)
65=b₁·(27+18+12+8)/27
b₁=27