Question

quantos termos tem a P.G.(5,20,80,...,1280)

Answer 1

PG

u1 = 5
u2 = 20

razão

q = u2/u1 = 20/5 = 4

termo geral
un = u1*q^(n - 1)

1280 = 5*4^(n - 1)

4^(n - 1) = 1280/5 = 256 = 4^4

n - 1 = 4

n = 5 termos