What is the 7th term of the geometric sequence where a1 = 256 and a3 = 16?

Well it is geometric. So we can know it is mtiplied or divided by something continuously. in this case divided because it is smaller. thus we can say the sequence is defined by:
a = 256 / x^(n - 1), where n is from 1 to infinite.

We have to solve for x to get the sequence and find the 7th term

so we know that we are looking at when n = 3, a = 16. so:
16 = 256/x^ (3- 1)
x^2 = 256/16
x^2 = 16
x = 4

so now we find 7th term
a = 256 / 4 ^(7- 1)
a = 256 / 4^6
a = 1 / 4^2
a = 1 / 16