Question

Determina o 13* termo da PG (64,32,16,..)

Answer 1

PG
a1 = 64
q = 32 : 64 = 1/2
n = 13
an = a13 = ?

an = a1 . $q^{n - 1}$
a13 = 64 . $1/2^{13 - 1}$
a13 = 64. $1/2^{12}$
a13 = 64 . 1/4.096
a13 = 64/4.096
a13 = 1/64

PG = (64, 32, 16, 8,  4, 2, 1, 1/2 , 1/4, 1/8 , 1/16, 1/32 , 1/64 )