SOLUTION: Find the sum to infinity of the following: 16, 12, 9, 27/4

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Question 1197507: Find the sum to infinity of the following: 16, 12, 9, 27/4
Found 2 solutions by math_helper, MathTherapy:
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
For a geometric series, the n-th partial sum is:
+S%5Bn%5D+=+a%5B1%5D%2A%28%281-r%5En%29%2F%281-r%29%29+

This problem presents a geometric sequence with a%5B1%5D+=+16 and r=3%2F4.
The infinite sum exists because abs%28r%29%3C1, and is the limit of S%5Bn%5D as n goes to infinity:
Using "Lim" to denote "Limit as n goes to infinity":
Lim (+S%5Bn%5D+) = Lim (+16%2A%281-%283%2F4%29%5En%29%2F%281-%283%2F4%29%29+ )

= Lim ( +16+%2A+4+%2A+%281-%283%2F4%29%5En%29+ )
= ( 16 * 4 * 1 )
= 64

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Find the sum to infinity of the following: 16, 12, 9, 27/4
This is a GEOMETRIC progression since it has a common ratio (r) of matrix%281%2C5%2C+12%2F16%2C+%22=%22%2C+9%2F12%2C+%22=%22%2C+3%2F4%29. Furthermore, |r| < 1.
  Sum of an INFINITE GEOMETRIC sequence: matrix%281%2C3%2C+S%5Binfinity%5D%2C+%22=%22%2C+a%5B1%5D%2F%281+-+r%29%29
                                         matrix%281%2C3%2C+S%5Binfinity%5D%2C+%22=%22%2C+16%2F%281+-+3%2F4%29%29 ----- Substituting 16 for a1, and 3%2F4 for r (common ratio) 
                                         matrix%281%2C3%2C+S%5Binfinity%5D%2C+%22=%22%2C+16%2F%281%2F4%29%29
                                              matrix%281%2C2%2C+%22=%22%2C+16%284%2F1%29%29

Sum of this INFINITE GEOMETRIC series: 16(4) = 64