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RemkaAna
@RemkaAna
July 2022
1
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Срочно!!!
Найдите первый член и разность арифметической прогрессии,у которой: A10=-30,S10=-120.
И ещё одно задание:
Сколько необходимо взять членов прогрессии 21;18;...,чтобы их сумма была равна нулю?
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nelle987
Verified answer
1)
S10 = (A1 + A10) * 10 / 2 = 5(A1 + A10)
A1 = S10 / 5 - A10 = -120 / 5 - (-30) = -24 + 30 = 6
d = (A10 - A1)/9 = (-30 - 6)/9 = -4
2) d = 18 - 21 = -3
Sn = (2a1 + d(n - 1))*n/2
(42 - 3(n - 1)) * n / 2 = 0
n не равно 0, можно на него делить
42 - 3(n - 1) = 0
14 - (n - 1) = 0
n = 15
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Answers & Comments
Verified answer
1)S10 = (A1 + A10) * 10 / 2 = 5(A1 + A10)
A1 = S10 / 5 - A10 = -120 / 5 - (-30) = -24 + 30 = 6
d = (A10 - A1)/9 = (-30 - 6)/9 = -4
2) d = 18 - 21 = -3
Sn = (2a1 + d(n - 1))*n/2
(42 - 3(n - 1)) * n / 2 = 0
n не равно 0, можно на него делить
42 - 3(n - 1) = 0
14 - (n - 1) = 0
n = 15