SOLUTION: The fourth, seventh and sixteenth terms of arithmetic series form a geometric series. If the first six terms of arithmetic series have a sum of 12. (a) Find the common difference

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Question 714016: The fourth, seventh and sixteenth terms of arithmetic series form a geometric series. If the first six terms of arithmetic series have a sum of 12.
(a) Find the common difference of the arithmetic series
(b) Calculate the common ratio of the geometric series

Answer by kevwill(135) About Me  (Show Source):
You can put this solution on YOUR website!
If x is the value of the first term, and d is the difference between terms, then the first 6 terms of the arithmetic series are:

x, x%2Bd, x%2B2d, x%2B3d, x%2B4d, x%2B5d

The sum of these 6 terms is 12, so we have:

6x+%2B+15d+=+12

The 4rd, 7th, and 16th terms, a geometric series, are

x%2B3d, x%2B6d, and x%2B15d

So we know that

%28x%2B6d%29%2F%28x%2B3d%29+=+%28x%2B15d%29%2F%28x%2B6d%29

Multiplying both sides by %28x%2B3d%29%2A%28x%2B6d%29

%28x%2B6d%29%2A%28x%2B6d%29+=+%28x%2B15d%29%2A%28x%2B3d%29

x%5E2+%2B+12xd+%2B+36d%5E2+=+x%5E2+%2B+18xd+%2B+45d%5E2

18xd+-+12xd+=+36d%5E2+-+45d%5E2

6xd+=+-9d%5E2

x+=+%28-3%2F2%29%2Ad where d%3C%3E0 (If d=0 we have the trivial sequence 2, 2, 2, 2, 2, ...)

Plugging x+=+%28-3%2F2%29d into 6x+%2B+15d+=+12

6%2A%28-3%2F2%29d+%2B+15d+=+12

-9d+%2B+15d+=+12

6d+=+12

d+=+2

And plugging d+=+2 into 6x+%2B+15d+=+12

6x+%2B+15%282%29+=+12

6x+%2B+30+=+12

6x+=+-18

x+=+-3

So our series is:

-3, -1, 1, 3, 5, 7, ...

The 4th, 7th, and 16th terms are 3, 9, and 27

So the difference in the arithmetic sequence is 2 and the ratio in the geometric sequence is 3.