Question 1132322: The numbers 3 and 192 are somewhere (but it is not known where) in a geometric sequence where all the terms in between are whole numbers. Find all the possibilities for the sequence.
I have found all 12 of the possibilities. They are:
2^6, 2^3, 2^2, 2^2, 2^1, 2^-6, 2^-3, 2^-2, 2^-1, -2^-2, -2^-1, -2^2 and -2^1
Solve the same problem with 5 and 3645. Explain why there are the same number of answers as
before.
I know the answer's for this question are the ones as before, but the Base for all of them are 3. So its:
3^6, 3^3, 3^2, 3^2, 3^1, 3^-6, 3^-3, 3^-2, 3^-1, -3^-2, -3^-1, -3^2 and -3^1
My question is that I need to know why they have the same amount of answers. If possible, I would like a step by step on how you got your answer. Thank you very much!
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
192 = 3*(2^6)
If 3 and 192 are terms of a sequence with all whole numbers, then the common ratio between terms has to be 2^n where n is a factor of 6 (if 192 is after 3 in the sequence) or 2^(-n) where n is a factor of 6 (if 192 is before 3 in the sequence.)
Those are the 12 answers you show for the possible common ratio.
3645 = 5*(3^6)
By the same reasoning as above, the possible common ratios are the 12 answers you show for this second sequence -- numbers of the form 3^n or 3^(-n) where n is a factor of 6.
If, for example, the numbers for the second sequence had been 5 and 405 = 5*(3^5), then there would have been only 4 possible common ratios -- numbers of the form 3^n or 3^(-n) where n is a factor of 5: 3^5, 3^1, 3^(-1), and 3^(-5).
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