2003 AIME I Problems/Problem 8
In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by . Find the sum of the four terms.
Denote the first term as , and the common difference between the first three terms as . The four numbers thus are in the form .
Since the first and fourth terms differ by , we have that . Multiplying out by the denominator, This simplifies to , which upon rearranging yields .
Both and are positive integers, so and must have the same sign. Try if they are both positive (notice if they are both negative, then and , which is a contradiction). Then, . Directly substituting and testing shows that , but that if then . Alternatively, note that or implies that , so only may work. Hence, the four terms are , which indeed fits the given conditions. Their sum is .
As another option, could be rewritten as follows:
This gives another way to prove , and when rewritten one last time:
shows that must contain a factor of 3.
EDIT by NealShrestha: Note that once we reach this implies since all other terms are congruent to .
The sequence is of the form . Since the first and last terms differ by 30, we have Let , where is an integer. This yields the following: We then set , where is an integer. Factoring using difference of squares, we have Then, noticing that , we set up several systems of equations involving the factors of . The second system we set up in this manner, yields the solution . Plugging back in, we get that , so the sequence is and the answer is
- Note: we do not have to check the other systems since the and values obtained via this system yield integers for , , and this must be the only possible answer since this is an AIME problem. We got very lucky in this sense :)
|2003 AIME I (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|