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rahat123
@rahat123
August 2022
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А3=-8; а5=4; S10-? Арифметическая прогрессия помогите пожалуйста
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IrkaShevko
Verified answer
S10 = (a1 + a10)*10/2 = 5(a1 + a1 + 9d) = 5(2a1 + 9d)
a3 + a5 = a1 + 2d + a1 + 4d = 2a1 +6d = -4
a1 + 3d = -2
a1 + 2d = -8
d = 6
a1 = -8 - 12 = -20
S10 = 5(-40 + 54) = 14*5 = 70
1 votes
Thanks 2
Munkush
И так воспользуемся формулой нахождения членов арифметической прогрессии
an=a1+d(n-1)
a1=an-d(n-1)
a1=a5-d(5-1) ; a5-4d=4-4*6=4-24=-20
a10=-20+54=34
а теперь формулой суммы арифмет прогрессии
1 votes
Thanks 1
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Answers & Comments
Verified answer
S10 = (a1 + a10)*10/2 = 5(a1 + a1 + 9d) = 5(2a1 + 9d)a3 + a5 = a1 + 2d + a1 + 4d = 2a1 +6d = -4
a1 + 3d = -2
a1 + 2d = -8
d = 6
a1 = -8 - 12 = -20
S10 = 5(-40 + 54) = 14*5 = 70
an=a1+d(n-1)
a1=an-d(n-1)
a1=a5-d(5-1) ; a5-4d=4-4*6=4-24=-20
a10=-20+54=34
а теперь формулой суммы арифмет прогрессии