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 :)
Solution 3 (Guesswork)
We represent the values as , , , and Take the difference between the first and last values Manipulating the values by expanding and then long division we see Combining like terms we get And looking at the values we know that since the second term must be positive (since both a and d are positive), must be a maximum of 9, which offers 9 possible values (since must be a positive integer.) We can resort to guesswork by this time, and thus receive the result of of at which . The solution is thus
- We could also lay down some further parameters for the value of . as must be greater than (for the first term to be positive), we can substitite in a "borderline" value of a=d (which itself is just over the limit). This gives us:
Thus, d must be greater than 7.5, which gives us only 2 values left (8 and 9). We can then plug in any of the two to see if an integer value is attainable. In the end, 8 doesn't work and 9 does, which gives the solution of
- Note: This solution is similar to Solution 2
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