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The sum of the first n terms in the infinite geometric sequence {1, 1/3, 1/9, 1/27, ...} is 121/81. Find n.

 

Edit: got it, it's 5 :D

AnonymousConfusedGuy  Jan 12, 2018
edited by AnonymousConfusedGuy  Jan 12, 2018
 #1
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1*[1 - (1/3)^N] / [1 - 1/3] =121/81, solve for N

 

3/2 (1 - 3^(-N)) = 121/81

 

Multiply both sides by 2/3:

1 - 3^(-N) = 242/243

 

Subtract 1 from both sides:

-3^(-N) = -1/243

 

Multiply both sides by -1:

3^(-N) = 1/243

 

Take reciprocals of both sides:

3^N = 243

 

243 = 3^5:

3^N = 3^5

 

Equate exponents of 3 on both sides:

N = 5

Guest Jan 12, 2018
edited by Guest  Jan 12, 2018

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