Question

6) quantos termos possui a pg (128,64,...,2)?

Answer 1

Fórmula: an = a1.q^n-1



an = 2
a1 = 128
n = ??
q = 1/2



2 = 128.(1/2)^n-1
2/128 = (1/2)^n-1
1/64 = (1/2)^n-1
(1/2)^6 = (1/2)^n-1
6 = n-1
6 + 1 = n
7 = n



A PG possui 7 termos.

Answer 2

Bom dia

PG(128, 64, ..., 2)

u1 = 128
u2 = 64

q = u2/u1 = 1/2

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

2 = 128*(1/2)^(n - 1)

1/64 = (1/2)^(n - 1)

2^(n - 1) =  64 = 2^6

n - 1 = 6

n = 7 termos