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gislainemartins1
@gislainemartins1
March 2022
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qual é o oitavo termo da pg(x3y7, x5y8,x7y9.....
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Geraldo5
Vamos obter a razão dessa PG. Sabemos que a razão é dada pela divisão de um número pelo seu antecessor, então:
q=x^5y^8/x^3y^7
q=x^2y
An=A1*(q)^(n-1)
A8=x^3y^7*(x^2y)^(8-1)
A8=x^3y*(x^2y)^7
A8=(x^3y)*(x^14y^7)
A8=x^17y^8
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Geraldo5
Qualquer erro, comunique
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q=x^5y^8/x^3y^7
q=x^2y
An=A1*(q)^(n-1)
A8=x^3y^7*(x^2y)^(8-1)
A8=x^3y*(x^2y)^7
A8=(x^3y)*(x^14y^7)
A8=x^17y^8