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flavioledson
@flavioledson
April 2022
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Insira três meios geométricos entre 3 e 768.
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cristianaplima
3, , , , 768.
a1=3
a5=768
q=?
a5=a1xq⁴
768=3x q⁴
768/3=q⁴
256=q⁴
q=⁴√264 q= ₄√4⁴ q=4
PG ( 3,12,48,192,768)
1 votes
Thanks 2
ashleybitt
colega vc pode me ajudar em uma questão?
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Calcule o sexto e o nono termos da PG ( 2,4,8,...).
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a1=3
a5=768
q=?
a5=a1xq⁴
768=3x q⁴
768/3=q⁴
256=q⁴
q=⁴√264 q= ₄√4⁴ q=4
PG ( 3,12,48,192,768)