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NatyRibeiro
@NatyRibeiro
December 2019
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A soma dos n primeiros termos da PG (3, -6, 12) é 129. Qual o valor de n ?
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albertrieben
Boa noite
PG
u1 = 3
u2 = -6
u3 = 12
q = u2/u1 = -6/3 = -2
soma
Sn = u1 * ( q^n - 1)/(q - 1)
3 * ( q^n - 1)/(q - 1) = 129
( q^n - 1)/(q - 1) = 43
(-2)^n - 1 = -129
(-2)^n = -128
(-2)^7 = -128
n = 7 termos
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PG
u1 = 3
u2 = -6
u3 = 12
q = u2/u1 = -6/3 = -2
soma
Sn = u1 * ( q^n - 1)/(q - 1)
3 * ( q^n - 1)/(q - 1) = 129
( q^n - 1)/(q - 1) = 43
(-2)^n - 1 = -129
(-2)^n = -128
(-2)^7 = -128
n = 7 termos