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>(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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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>.
  
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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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>.
  
 
== See also ==
 
== See also ==

Revision as of 17:33, 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 $a$, and the common difference between the first three terms as $d$. The four numbers thus resemble $\displaystyle a,\ a+d,\ a+2d,\ \frac{(a + 2d)^2}{a + d}$.

Since the first and fourth terms differ by 30, we can write that $\frac{(a + 2d)^2}{a + d} - a = 30$. Multiplying out by the denominator, we get that $\displaystyle (a + 4ad + 4d^2) - a(a + d) = 30(a + d)$. This simplifies to $\displaystyle 3ad + 4d^2 = 30a + 30d$. Rearrange the terms to find that $\displaystyle 2d(2d - 15) = 3a(10 - d)$.

Both $a$ and $d$ are positive integers, so $2d - 15$ and $10 - d$ must have the same sign. Try if they are both positive (notice if they are both negative, then $\displaystyle d > 10$ and $d < \frac{15}{2}$, which clearly is a contradiction). Then, $d = 8, 9$. Directly substituting and testing shows that $\displaystyle d \neq 8$, but that if $d = 9$ then $a = 18$. Hence, the four terms are $18,\ 27,\ 36,\ 48$, which indeed fits the given conditions. Their sum is $129$.

See also

2003 AIME I (ProblemsAnswer KeyResources)
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