SOLUTION: 2) Use the geometric sequence of numbers 1, 3, 9, 27, … to find the following: a) What is r, the ratio between 2 consecutive terms? Answer: Show work in this space

Algebra ->  Sequences-and-series -> SOLUTION: 2) Use the geometric sequence of numbers 1, 3, 9, 27, … to find the following: a) What is r, the ratio between 2 consecutive terms? Answer: Show work in this space      Log On


   



Question 78546: 2) Use the geometric sequence of numbers 1, 3, 9, 27, … to find the following:
a) What is r, the ratio between 2 consecutive terms?
Answer:
Show work in this space.



b) Using the formula for the nth term of a geometric sequence, what is the 10th term?
Answer:
Show work in this space.



c) Using the formula for the sum of a geometric sequence, what is the sum of the first 10 terms?
Answer:
Show work in this space.

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 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=%283%29%2F%281%29=3I can also do it with 27 and 9 and it will still give me the same value
r=%2837%29%2F%289%29=3
So r=3


b) Since we are multiplying by r each term, our sequence is simply

a%5Bn%5D=a%5Bn-1%5D%2Ar

So to find the nth term (a%5Bn%5D), we simply multiply the previous term (a%5Bn-1%5D) by r.

So if we start at 1, to get to 3 we multiply 1 by r=3

3=1%2A3

Now to go from 3 to 9 we multiply by r again

9=1%2A3%2A3

Now to go from 9 to 27 we multiply by r again

27=1%2A3%2A3%2A3

Notice that for the 1st term 3 we have only 1 r, 2nd term we have 2 r's, etc. So the term we have determines the number of r's. In other words, the nth term is

a%5Bn%5D=1%2Ar%5En

So our sequence is a%5Bn%5D=1%2Ar%5En

Now let n=9 to find the 10th term (we started at n=0)

a%5Bn%5D=1%2A3%5E9=19683

So the 10th term is 19,683


c) 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=3
s=%281-%283%29%5E10%29%2F%281-%283%29%29
s=%281-59049%29%2F%28-2%29
s=29524
So the sum of the first ten terms are 29,524