SOLUTION: Find three numbers in a geometric sequence where the third term is 4 times the first term and the difference between the second and the third term is 24.

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Question 1119641: Find three numbers in a geometric sequence where the third term is 4 times the first term and the difference between the second and the third term is 24.
Found 2 solutions by solver91311, ikleyn:
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!


The term of a geometric series is:



Where is the common ratio:

So the first three terms are or more simply:



Since is given to be , we know that



And since





So the first three terms are:




John

My calculator said it, I believe it, that settles it
Answer by ikleyn(52795)   (Show Source): You can put this solution on YOUR website!
.
The problem has two solutions  (two answers).

One answer is the sequence  12, 24, 48,  found by John.

The second answer is the sequence  4, -8, 16.

Solution 

The three terms are  a, ar and ar^2.


From the first part of the condition ("the third term is 4 times the first term") you have

ar^2 = 4a,


which implies  r^2 = 4  and, hence,  r = +/- 2.


1)  Let r = 2.  Then you get the sequence  12, 24, 48,  as John obtained it.


2)  Let r = -2.  


    Then the second part of the condition says

    ar^2 - ar = 24  ====>  a*4 - a*(-2) = 24  ====>  6a = 24  ====>  a = 4,


    and you get the second sequence  4, -8, 16.

Solved.


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