SOLUTION: Is this correct? Use the geometric sequence of numbers 1, 1/2, 1/4, 1/8,…to find the following: What is r, the ratio between 2 consecutive terms? Answer: r = (1/2)/1 = 1/2

Algebra ->  Algebra  -> Sequences-and-series -> SOLUTION: Is this correct? Use the geometric sequence of numbers 1, 1/2, 1/4, 1/8,…to find the following: What is r, the ratio between 2 consecutive terms? Answer: r = (1/2)/1 = 1/2      Log On

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Question 83996: Is this correct?
Use the geometric sequence of numbers 1, 1/2, 1/4, 1/8,…to find the following:
What is r, the ratio between 2 consecutive terms?
Answer: r = (1/2)/1 = 1/2
Show work in this space.
Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 10 terms? Please round your answer to 4 decimals.
Answer:
Show work in this space.
S(n) = a(1)[r ^(n + 1) – 1)/(r – 1)
S(10) = 1[(1/2) ^ 9 – 1] / [(1/2)-1]
S(10) = [-0.998046875…] / [-0.5] = 1.99609375…
Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 12 terms? Please round your answer to 4 decimals.
Answer:
Show work in this space.
S(n) = a(1)[r ^(n + 1) – 1)/(r – 1)
S(12) = 1[(1/2) ^ 9 – 1] / [(1/2)-1]
S(12) = [-0.998046875…] / [-0.5] = 1.99609375…
n = 12

Answer by jim_thompson5910(14933) About Me  (Show Source):
You can put this solution on YOUR website!

The ratio r is the factor to get from term to term. So
r=nth term/(n-1) term
r=%281%2F8%29%2F%281%2F4%29=%284%2F8%29=1%2F2
r=1%2F2
The sequence is cut in half each term, so the sequence is %281%2F2%29%5En


The sum of a geometric series is
S=a%281-r%5En%29%2F%281-r%29where a=1
S=%281-%281%2F2%29%5E10%29%2F%281-%281%2F2%29%29So plug in n=10 to find the sum of the first 10 partial sums
S=%281-1%2F1024%29%2F%281%2F2%29
S=2046%2F1024
So the sum of the first ten terms is 2046%2F1024 or 1.99805 approximately

Use the same formula to find the sum of the 1st 12 terms
S=a%281-r%5En%29%2F%281-r%29where a=1
S=%281-%281%2F2%29%5E12%29%2F%281-%281%2F2%29%29So plug in n=12 to find the sum of the first 12 partial sums
S=%281-1%2F4096%29%2F%281%2F2%29
S=8190%2F4096
So the sum of the first twelve terms is 8190%2F4096 or 1.99951 approximately