First term of an infinite geometric sequence

  • Thread starter thornluke
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  • #1
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Homework Statement


The sum of an infinite geometric sequence is 131/2, and the sum of the first three terms is 13. Find the first term.


Homework Equations


S = a/(1-r)
Sn = a-arn/(1-r)


The Attempt at a Solution


a/(1-r) = 131/2

a-ar3/(1-r) = 13

2a = 27-27r ........................ 1
a-ar3 = 13-13r..... 2

I'm stuck.
 

Answers and Replies

  • #2
205
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[itex]2a=27-27r[/itex] (1)
[itex]a-ar^3=13-13r[/itex] (2)

Put 27 in evidence in (1) and put 13 in evidence in (2).

[itex]2a=27(1-r)[/itex] (1)
[itex]a-ar^3=13(1-r)[/itex] (2)

Now divide (2) by (1).
 
  • #3
HallsofIvy
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Oh!! It is the "a" that cancels allowing you to first solve for r. I was too focused on finding a. Very good pc2-brazil!
 
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  • #4
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[itex]2a=27-27r[/itex] (1)
[itex]a-ar^3=13-13r[/itex] (2)

Put 27 in evidence in (1) and put 13 in evidence in (2).

[itex]2a=27(1-r)[/itex] (1)
[itex]a-ar^3=13(1-r)[/itex] (2)

Now divide (2) by (1).
[itex]2a=27-27r[/itex] (1)
[itex]a-ar^3=13-13r[/itex] (2)

Put 27 in evidence in (1) and put 13 in evidence in (2).

[itex]2a=27(1-r)[/itex] (1)
[itex]a-ar^3=13(1-r)[/itex] (2)

Now divide (2) by (1).
I found that a =9.
 
  • #5
205
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I found that a =9.
That is correct. I suppose you also found that r = 1/3.
Then, the sum of the first three terms is 13:
9 + 3 + 1 = 13
The sum of the infinite geometric sequence is 13.5:
9 + 3 + 1 + 1/3 + ... = [itex]\frac{9}{1-\frac{1}{3}}=13.5[/itex]
 

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