Question

caucule quantos termos tem a pg,
an= 390625 a1=125 e q=25

Answer 1

an = a1.q^(n-1)
390625 = 125.25^(n - 1)
390625 = 5^3 . 5^(2n-2)
390625 = 5^(3+2n-2)
390625= 5^(1+2n)
5^8 = 5^(1+2n)
8 = 1 + 2n
8 - 1 = 2n
7 = 2n
2n = 7
n = 7/2
n = 3,5