2003 AIME I Problems/Problem 8
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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,
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
.
Postscript
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.
-jackshi2006
EDIT by NealShrestha:
Note that once we reach this implies
since all other terms are congruent to
.
Solution 2
The sequence is of the form
. Since the first and last terms differ by 30, we have
Let
, where
is an integer. This yields
$$ (Error compiling LaTeX. Unknown error_msg)a=\frac{-120 \pm
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 |
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