Difference between revisions of "2003 AIME I Problems/Problem 8"
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== Solution == | == Solution == | ||
| − | {{ | + | Denote the first term as <math>a</math>, and the common difference between the first three terms as <math>d</math>. The four numbers thus resemble <math>\displaystyle a,\ a+d,\ a+2d,\ \frac{(a + 2d)^2}{a + d}</math>. |
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| + | Since the first and fourth terms differ by 30, we can write that <math>\frac{(a + 2d)^2}{a + d} - a = 30</math>. Multiplying out by the [[denominator]], we get that <math>(a + 4ad + 4d^2) - a(a + d) = 30(a + d)</math>. This simplifies to <math>3ad + 4d^2 = 30a + 30d</math>. Rearrange the terms to find that <math>2d(2d - 15) = 3a(10 - d)</math>. | ||
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| + | Both <math>a</math> and <math>d</math> are positive integers, so <math>2d - 15</math> and <math>10 - d</math> must have the same sign. Try if they are both [[positive]] (notice if they are both [[negative]], then <math>d > 10</math> and <math>d < \frac{15}{2}</math>, which clearly is a contradiction). Then, <math>d = 8, 9</math>. Directly substituting and testing shows that <math>d \neq 8</math>, but that if <math>d = 9</math> then <math>a = 18</math>. Hence, the four terms are <math>18,\ 27,\ 36,\ 48</math>, which indeed fits the given conditions. Their sum is <math>129</math>. | ||
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== See also == | == See also == | ||
| − | + | {{AIME box|year=2003|n=I|num-b=7|num-a=9}} | |
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 16:29, 8 March 2007
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 | ||
