What is the sixth term in the geometric sequence $\frac{27}{125}, \frac{9}{25}, \frac{3}{5},\ldots$? Express your answer as a common fraction.
\( $\frac{27}{125}, \frac{9}{25}, \frac{3}{5},\ldots$\)
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
