Question

Determine o 15° terno da PG (25 6,128,64,31...)

Answer 1

PG u1 = 256  u2 = 128  razão  q = u2/u1 = 128/256 = 1/2  un = u1*q^(n - 1)  u15 = 256*(1/2)^14  u15 = 2^8/2^14 = 2^-6 = 1/64  pronto

Answer 2

(256,128,64,32...)
a1=256
an=?
q=1/2
n=15
An=a1,q^n-1
An=256.(1/2)^15-1
An=256.(1/2)^14
An=256.(1/2)
An=256/16.384
An=256:256/16.384:256
An=1/64