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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.

Guest Oct 2, 2017
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\( $\frac{27}{125}, \frac{9}{25}, \frac{3}{5},\ldots$\)

 

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

 

 

cool cool cool

CPhill  Oct 2, 2017

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