SOLUTION: The sum to infinity of a geometric series is 8, whereas the sum of the second and third terms is 3. What are all the possible values of the common ratio of the series?

Algebra ->  Sequences-and-series -> SOLUTION: The sum to infinity of a geometric series is 8, whereas the sum of the second and third terms is 3. What are all the possible values of the common ratio of the series?      Log On


   

Question 1191284: The sum to infinity of a geometric series is 8, whereas the sum of the second and third terms is 3. What are all the possible values of the common ratio of the series?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Let a and r be the first term and common ratio. The given information tells us

(1) a+ar+ar^2+ar^3+...=8
(2) ar+ar^2=3

We can write the infinite sum using the formula for the sum of an infinite geometric series:

(3) a%2F%281-r%29=8

Subtracting (2) from (1), we get

(4) a+ar^3+ar^4+...=5

Use the formula for the sum of an infinite geometric series again on all but the first term:

(5) a%2B%28ar%5E3%2F%281-r%29%29=5

Substitute (3) into (5):

a%2B8r%5E3=5
8r%5E3=5-a
r%5E3=%285-a%29%2F8
(6) r=%28%285-a%29%2F8%29%5E%281%2F3%29

Substitute (6) into (3):

a%2F%281-%28%285-a%29%2F8%29%5E%281%2F3%29%29=8


4%285-a%29%5E%281%2F3%29=8-a
%285-a%29%5E%281%2F3%29=2-a%2F4

At this point, it is easily seen that a=4 will work.

Algebraic or numerical methods, or a graphing calculator, can be used to show that a=4 is the ONLY solution to that equation, so it will lead us to the unique answer for the common ratio r.

r=%28%285-a%29%2F8%29%5E%281%2F3%29=%281%2F8%29%5E%281%2F3%29=1%2F2

ANSWER: the common ratio of the series is 1/2

CHECK:

The sequence has first term 4 and common ratio 1/2: 4, 2, 1, 1/2, 1/4, ...
The infinite sum is 4%2F%281-1%2F2%29=8
The sum of the second and third terms is 2+1=3

NOTE:
The solution a=4 and r=1/2 can be found by playing with numbers on the original problem. Since the problem asked for the sum of all the possible values of the common ratio, we had to do some algebra to show that there are no other possible values for the common ratio.