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

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) (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:

, , , , ,

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

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

, , and

So we know that

Multiplying both sides by

where (If we have the trivial sequence , , , , , ...)

Plugging into

And plugging into

So our series is:

, , , , , , ...

The 4th, 7th, and 16th terms are , , and

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

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