Let r be the common ratio
Let the first term be a, the second term, ar, the fourth term,ar^3 and the fifth term, ar^4
So we have
a + ar^4 = 1285 → ar^4 = 1285 - a (1)
ar * ar^3 = 6400 → a (ar^4) = 6400 (2)
Subbing (1) into (2) we have
a (1285 - a) = 6400
1285a - a^2 = 6400 rearrange as
a^2 - 1285a + 6400 = 0 factor
(a - 1280) (a - 5) = 0
So....setting each factor to 0 and solving for a produces a = 1280 or a = 5
If a = 1280, we have
1280r^4 = 1285 - 1280
1280r^4 = 5
r^4 = 5 /1280 = 1 /256 → r = 1/4
So....one possible series is
1280, 320, 80, 20, 5
And if a = 5, we have
5r^4 = 1285 - 5
5r^4 = 1280
r^4 = 256 → r = 4
And another possible series is
5, 20, 80, 320, 1280