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An infinite geometric series has first term $328$ and a sum of $2009$. What is its common ratio?

 
michaelcai  Nov 14, 2017

Best Answer 

 #2
avatar
+1

S = F / [1 - R] 

2009 = 328 / [1 - R]

 

Solve for R:
328/(1 - R) = 2009

Take the reciprocal of both sides:
(1 - R)/328 = 1/2009

Multiply both sides by 328:
1 - R = 8/49

Subtract 1 from both sides:
-R = -41/49

Multiply both sides by -1:
R = 41/49

 
Guest Nov 14, 2017
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2+0 Answers

 #1
avatar+1961 
+2

2009 = 328/(1 - r)

 

Solve for r, and there you go! :)

This is using the formula:

S = term1/(1 - r)

 
saseflower  Nov 14, 2017
 #2
avatar
+1
Best Answer

S = F / [1 - R] 

2009 = 328 / [1 - R]

 

Solve for R:
328/(1 - R) = 2009

Take the reciprocal of both sides:
(1 - R)/328 = 1/2009

Multiply both sides by 328:
1 - R = 8/49

Subtract 1 from both sides:
-R = -41/49

Multiply both sides by -1:
R = 41/49

 
Guest Nov 14, 2017

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