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Find the common ratio of the infinite geometric series: $$\frac{-3}{5}-\frac{5}{3}-\frac{125}{27}-\dots$$

 Jan 28, 2015

Best Answer 

 #2
avatar+118609 
+5

The common ratio is given by

$$\\r=T_{n+1}\div T_{n}\\\\
so \\
r= \frac{-5}{3}\div\frac{-3}{5}\\\\
r= \frac{-5}{3}\times\frac{-5}{3}\\\\
r= \frac{25}{9}\\\\\\
OR\\
r= \frac{-125}{27}\div\frac{-5}{3}\\\\
r= \frac{-125}{27}\times\frac{3}{-5}\\\\
r= \frac{25}{9}\times\frac{1}{1}\\\\
r= \frac{25}{9}$$

 

IN FACT ANY TERM DIVIDED BY THE TERM BEFORE IT WILL GIVE THE COMMON RATIO    

 Jan 28, 2015
 #1
avatar+128475 
+5

The common ratio is 25/9  

 

 

 Jan 28, 2015
 #2
avatar+118609 
+5
Best Answer

The common ratio is given by

$$\\r=T_{n+1}\div T_{n}\\\\
so \\
r= \frac{-5}{3}\div\frac{-3}{5}\\\\
r= \frac{-5}{3}\times\frac{-5}{3}\\\\
r= \frac{25}{9}\\\\\\
OR\\
r= \frac{-125}{27}\div\frac{-5}{3}\\\\
r= \frac{-125}{27}\times\frac{3}{-5}\\\\
r= \frac{25}{9}\times\frac{1}{1}\\\\
r= \frac{25}{9}$$

 

IN FACT ANY TERM DIVIDED BY THE TERM BEFORE IT WILL GIVE THE COMMON RATIO    

Melody Jan 28, 2015

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