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?

One way to do this is to note that the second and fourth numbers both have to be perfect squares times a number. (the same number) So start out with the perfect squares for the second term and fourth term.

1 won't work as the second term or fourth term, because no integer below it is positive.

Trying 4 as the second term and 9 as the fourth term: 2 4 6 9. The difference between 2 and 9 is 7, and 7 is not a factor of 30.

Trying 4 as the second term and anything more than 9 for the fourth term will end up with the first term not being positive.

So trying 9 as the second term and 16 as the fourth term: 6 9 12 16, and this will work because the difference between the first term and fourth term is 10, a factor of 30. So multiplying every term by 30/10 or 3 gives 18, 27, 36, 48, and 48 - 18 is 30 as the problem says, so 18, 27, 36, 48 is the solution.

Other possible sequences include:
3, 9, 15, 25, and 25 - 3 = 22
12, 16, 20, 25, and 25 - 12 = 13
8, 16, 24, 36, and 36 - 8 = 28
20, 25, 30, 36, and 36 - 20 = 16
30, 36, 42, 49, and 49 - 30 = 19
42, 49, 56, 64, and 64 - 42 = 22
56, 64, 72, 81, and 81 - 56 = 25
72, 81, 90, 100, and 100 - 72 = 28

All other combinations won't work because they will end up with first terms which aren't positive, or the difference between the first and last term will be greater than 30.

None of these sequences have the difference between the first and last term being a factor of 30, so no other sequence is possible.

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