SOLUTION: Use the geometric sequence of numbers 1, 1/3, 1/9 , 1/27… to find the following: a) What is r, the ratio between 2 consecutive terms? b) Using the formula for the sum of t

Algebra ->  Sequences-and-series -> SOLUTION: Use the geometric sequence of numbers 1, 1/3, 1/9 , 1/27… to find the following: a) What is r, the ratio between 2 consecutive terms? b) Using the formula for the sum of t      Log On


   

Question 78942: Use the geometric sequence of numbers 1, 1/3, 1/9 , 1/27… to find the following:
a) What is r, the ratio between 2 consecutive terms?
b) Using the formula for the sum of the first n terms of a geometric sequence, what is the sum of the first 10 terms? Carry all calculations to 6 decimals on all assignments.
c) Using the formula for the sum of the first n terms of a geometric sequence, what is the sum of the first 12 terms? Carry all calculations to 6 decimals on all assignments.
d) What observation can make about the successive partial sums of this sequence? In particular, what number does it appear that the sum will always be smaller than?

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a)
The ratio r is the factor needed to go from term to term. To find the factor, divide any term by its previous term. So I chose 1/3 as the first term to be divided by 1
Ratio r: r=%28nth+term%29%2F%28n-1+term%29pick any nth term and any previous term, such as the 2nd and 1st term.
r=%281%2F3%29%2F%281%29=1%2F3I can also do it with 1/27 and 1/9 and it will still give me the same value
r=%281%2F27%29%2F%281%2F9%29=%289%2F27%29=%281%2F3%29
So r=1/3
b) The sum of the first ten terms can be found by using
s=%281-r%5E%28n%2B1%29%29%2F%281-r%29
So let r=1/3
s=%281-%281%2F3%29%5E10%29%2F%281-%281%2F3%29%29
s=%281-%281%2F59049%29%29%2F%282%2F3%29
s=1.499975
c) Use the same technique but with 12 terms
s=%281-r%5E%28n%2B1%29%29%2F%281-r%29
let r=1/3
s=%281-%281%2F3%29%5E12%29%2F%281-%281%2F3%29%29
s=%281-%281%2F531441%29%29%2F%282%2F3%29
s=1.4999971 If it seems this sum is approaching 1.5 it is. If you go to the millionth term it will be even closer to 1.5
d)Using what we found from the two previous problems we can see that the partial sums, the pieces that make up the whole sum, are smaller than its previous term. This allows the entire series to approach 1.5