Question

What is the 8th term of the geometric sequence where a1 = 256 and a3 = 16?

Answer 1

common ratio: r^2 = 16 / 256 = 1 /16 so r = 1/4
so 8th term = a* (1/4)^7 = 256 / (1/4)^7

Answer 2

tn=ar^(n-1)

t1=256*r^(1-1)
t3=256*r^2
16=256*r^2
1/16=r^2
r=sqrt(1/16)
r=1/4

t8=256*(1/4)^7
t8=1/64

Answer 3

yup