?????

$$\frac{27}{125}, \frac{9}{25}, \frac{3}{5},\ldots$$

The series is 

27          (1/3)^(n - 1)

____  *                             where n  is the nth term

125        (1/5)^(n - 1)

So.......the 6th  term  is

27 * (1/3)^5 /  [125 * (1/5)^5]  =

(27 / 243) / [ 125 /3125]  =

(1/9) / (1/25)  =

25/9