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

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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

#2
+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
Sort:

#1
+1965
+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
+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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