SOLUTION: the sum of the first six terms of a geometric progression is nine time the sum of the first three terms, where neither the first nor the common ratio is equal to zero. find the com

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Question 277454: the sum of the first six terms of a geometric progression is nine time the sum of the first three terms, where neither the first nor the common ratio is equal to zero. find the common ratio.
Answer by Edwin McCravy(20086) About Me  (Show Source):
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the sum of the first six terms of a geometric progression is nine time the sum of the first three terms, where neither the first nor the common ratio is equal to zero. find the common ratio.


Let the first term be a and the common ratio be r

Sum of first three terms = a%2Bar%2Bar%5E2

Let this sum of the first three terms be X

Sum of first six terms = a%2Bar%2Bar%5E2%2Bar%5E3%2Bar%5E4%2Bar%5E5

Sum of first six terms = X%2Bar%5E3%2Bar%5E4%2Bar%5E5=X%2Br%5E3%28a%2Bar%2Bar%5E2%29=X%2Br%5E3X

We are given that X%2Br%5E3X=9X.  Divide through by X and we have

1%2Br%5E3=9

r%5E3=8

r=root%283%2C8%29

r=2

Check: Suppose the first term is a. Then the first six terms are

a,2a,4a,8a,16a,32a

The sum of the first three terms is a+2a+4a = 7a

The sum of the first 6 terms is a+2a+4a+8a+16a+32a=63a which is 9 times 7a.

So the answer is r=2.

Edwin