SOLUTION: Let a + ar + ar^2 + ar^3 + ... be an infinite geometric series. The sum of the series is 3. The sum of the cubes of all the terms is 5. Find the common ratio.

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Question 1209387: Let
a + ar + ar^2 + ar^3 + ...
be an infinite geometric series. The sum of the series is 3. The sum of the cubes of all the terms is 5. Find the common ratio.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The given series is

a+ar+ar^2+ar^3+...

The sum of the series is 3:

[1] a%2F%281-r%29=3

The series consisting of the cubes of the terms of the given series is

a^3r^3+a^3r^6+a^3r^9+...

The sum of that series is 5:

[2] a%5E3%2F%281-r%5E3%29=5

To find the common ratio r, solve [1] for a in terms of r and substitute in [2].

a=3%281-r%29

%283%281-r%29%29%5E3%2F%281-r%5E3%29=5

27%281-3r%2B3r%5E2-r%5E3%29=5%281-r%5E3%29

27-81r%2B81r%5E2-27r%5E3=5-5r%5E3

22r%5E3-81r%5E2%2B81r-22=0

By inspection, r=1 is one solution to that equation. However r=1 produces an infinite geometric series that has no sum.

Use synthetic division to remove the root x=1 to find the other two roots.

  1 | 22 -81  81 -22
    |     22 -59  22
    +---------------
      22 -59  22   0

The remaining quadratic is 22x%5E2-59x%2B22

Use the quadratic formula to find that the other two solutions are

%2859%2B-sqrt%281545%29%29%2F44

There are two possible values for the common ratio r:

ANSWERS:
%281%2F44%29%2859%2Bsqrt%281545%29%29
%281%2F44%29%2859-sqrt%281545%29%29