2003 AIME I Problems/Problem 8
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 30. 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 resemble
.
Since the first and fourth terms differ by 30, we can write that . Multiplying out by the denominator, we get that
. This simplifies to
. Rearrange the terms to find that
.
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 clearly is a contradiction). Then,
. Directly substituting and testing shows that
, but that if
then
. 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 |