Question

(08.03) what is the 8th term of the geometric sequence where a1 = 256 and a3 = 16?

Answer 1

a1 = a = 256
a3 = ar^2 = 16

256×r^2 = 16
r^2 = 1/16
hence r= 1/4

a8 = ar^7 = 256×(1/4)^7
=256/16384

= 64