Difference between revisions of "2003 AIME I Problems/Problem 8"
(minor) |
|||
| Line 15: | Line 15: | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 18:58, 4 July 2013
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.
Solution
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,
![\[(a^2 + 4ad + 4d^2) - a(a + d) = 30(a + d).\]](http://latex.artofproblemsolving.com/0/7/f/07fd58d0963b72126bae86ec79dbb742881168b1.png)
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 
.
See also
| 2003 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
