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? Answer: Show work in this space.

Algebra ->  Quadratic Equations and Parabolas -> 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? Answer: Show work in this space.       Log On


   

Question 87238: 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?
Answer:
Show work in this space.



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.
Answer:
Show work in this space.



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.
Answer:
Show work in this space.



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:


Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a)
The ratio r is the factor to get from term to term. So to find r, simply pick any term and divide it by the previous term:
r=%281%2F27%29%2F%281%2F9%29=%289%2F27%29=1%2F3 Divide the term of 1%2F27 by 1%2F9
So the ratio is
r=1%2F3
The sequence is reduced by a factor of 1%2F3 each term, so the sequence is %281%2F3%29%5En

b)
The sum of a geometric series is
S=a%281-r%5En%29%2F%281-r%29where a=1

S=%281-%281%2F3%29%5E10%29%2F%281-%281%2F3%29%29 So plug in r=1%2F3 and n=10 to find the sum of the first 10 partial sums

S=%281-1%2F59049%29%2F%281-%281%2F3%29%29 Raise 1%2F3 to the 10 power

S=%2859049%2F59049-1%2F59049%29%2F%281%2F2%29 Make "1" into an equivalent fraction with a denominator of 59049

S=%2859048%2F59049%29%2F%281-%281%2F3%29%29 Subtract the fractions in the numerator

S=%2859048%2F59049%29%2F%282%2F3%29 Subtract the fractions in the denominator

S=177144%2F118098 Flip the 2nd fraction, multiply, and simplify

So the sum of the first ten terms is 177144%2F118098 or 1.49997459736829 approximately

note: I chose to use fractions (to maintain accuracy), but it may be much easier for you to simply use a calculator to evaluate the sum.
c)
The sum of a geometric series is

S=a%281-r%5En%29%2F%281-r%29where a=1

S=%281-%281%2F3%29%5E12%29%2F%281-%281%2F3%29%29 So plug in r=1%2F3 and n=12 to find the sum of the first 12 partial sums

S=%281-1%2F531441%29%2F%281-%281%2F3%29%29 Raise 1%2F3 to the 10 power

S=%28531441%2F59049-1%2F59049%29%2F%281%2F2%29 Make "1" into an equivalent fraction with a denominator of 531441

S=%28531440%2F531441%29%2F%281-%281%2F3%29%29 Subtract the fractions in the numerator

S=%28531440%2F531441%29%2F%282%2F3%29 Subtract the fractions in the denominator

S=1594320%2F1062882 Flip the 2nd fraction, multiply, and simplify

So the sum of the first twelve terms is 1594320%2F1062882 or 1.49999717748537 approximately

d)
It appears that the sums are approaching a finite number of 3%2F2 or 1.5. This is because each term is getting smaller and smaller. This observation is justified by the fact that if abs%28r%29%3C1 then the infinite series will approach a finite number. In other words
If abs%28r%29%3C1 (the magnitude of r has to be less than 1) then,
S=a%2F%281-r%29Where S is the sum of the infinite series. So if we let a=1 and r=1/3 we get
S=1%2F%281-%281%2F3%29%29
S=1%2F%282%2F3%29
S=3%2F2So this verifies that our series approaches 3%2F2.