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MaluPoliciano
@MaluPoliciano
January 2020
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A soma dos N primeiros termos da progressão aritmética (6,10,14...) é 510.Calcular n.
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LouisXV
Ola
PA
a1 = 6
a2 = 10
a3 = 14
r = a2 - a1 = 10 - 6 = 4
soma
Sn = a1*n + r*(n - 1)*n/2 = 510
Sn = 6n + 4*(n - 1)*n/2 = 510
6n + 2n² - 2n - 510 = 0
2n² + 4n - 510 = 0
n² + 2n - 255 = 0
delta
d² = 4 + 4*255 = 4*256
d = 2*16 = 32
n = (-2 + 32)/2 = 30/2 = 15 termos
4 votes
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MaluPoliciano
mto obg!!!
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Lista de comentários
PA
a1 = 6
a2 = 10
a3 = 14
r = a2 - a1 = 10 - 6 = 4
soma
Sn = a1*n + r*(n - 1)*n/2 = 510
Sn = 6n + 4*(n - 1)*n/2 = 510
6n + 2n² - 2n - 510 = 0
2n² + 4n - 510 = 0
n² + 2n - 255 = 0
delta
d² = 4 + 4*255 = 4*256
d = 2*16 = 32
n = (-2 + 32)/2 = 30/2 = 15 termos