Difference between revisions of "2003 AIME I Problems/Problem 8"
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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>. | 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>. | ||
− | 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>\displaystyle (a + 4ad + 4d^2) - a(a + d) = 30(a + d)</math>. This simplifies to <math>\displaystyle 3ad + 4d^2 = 30a + 30d</math>. Rearrange the terms to find that <math>\displaystyle 2d(2d - 15) = 3a(10 - d)</math>. | + | 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>\displaystyle (a^2 + 4ad + 4d^2) - a(a + d) = 30(a + d)</math>. This simplifies to <math>\displaystyle 3ad + 4d^2 = 30a + 30d</math>. Rearrange the terms to find that <math>\displaystyle 2d(2d - 15) = 3a(10 - d)</math>. |
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>\displaystyle 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>\displaystyle 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>. | 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>\displaystyle 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>\displaystyle 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>. |
Revision as of 15:24, 26 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 |