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Sequence Group (Posted on 2003-09-23) Difficulty: 3 of 5
I didn't come up with this problem, but I still think it's a good one.

There are 4 positive integers in order from least to greatest, such that the first three make an aritmetic sequence, and the last three make a geometric sequence. If the difference between the largest and smallest term is 30, what are the terms?

See The Solution Submitted by Gamer    
Rating: 3.4000 (5 votes)

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Solution solution | Comment 2 of 8 |
If w is the first integer and a is the difference, then the first three terms are
w, w+a, w+2a.

In order for the last three terms to be in geometric sequence, the ratio of the last (4th) term to the 3rd must be the same as that of the 3rd to the 2nd:


so the last term must be


so that the difference between the last term and the first will be

(w+2a)^2/(w+a) - w

which must be set to 30. After that is done we can multiply both sides by (w+a), giving

(w+2a)^2 - w^2 - wa = 30 (w+a)
w^2+4wa+4a^2-w^2-wa = 30w + 30a
w(3a-30) = -4a^2 + 30a
w = (4a^2 - 30a)/(30-3a)

The only positive integral value for a which gives a positive integral value for w is 9, which gives w=18, so the sequence is

18, 27, 36, 48
  Posted by Charlie on 2003-09-23 20:55:22
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