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lena2602
@lena2602
September 2021
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Помогите))))
Выписаны первые несколько членов геометрической прогрессии: −750; 150; −30; … Найдите сумму первых 5 её членов.
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nafanya2014
Verified answer
B₁ = - 750; b₂ = 150; b₃ = - 30.
q=b₂:b₁=150:(-750)=-1/5
q=b₃:b₂=-30:150=-1/5
b₄=b₃·q=(-30)·(-1/5)=6
b₅=b₄·q=6·(-1/5)=-6/5
S = b₁+b₂+b₃+b₄+b₅ = -750 + 150 + (-30) + 6 +(-6/5) = -625 целых 1/5 =
= - 625,2
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Verified answer
B₁ = - 750; b₂ = 150; b₃ = - 30.q=b₂:b₁=150:(-750)=-1/5
q=b₃:b₂=-30:150=-1/5
b₄=b₃·q=(-30)·(-1/5)=6
b₅=b₄·q=6·(-1/5)=-6/5
S = b₁+b₂+b₃+b₄+b₅ = -750 + 150 + (-30) + 6 +(-6/5) = -625 целых 1/5 =
= - 625,2