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 $a$ , and the common difference between the first three terms as $d$ . The four numbers thus are in the form $a,\ a+d,\ a+2d,\ \frac{(a + 2d)^2}{a + d}$ .

Since the first and fourth terms differ by $30$ , we have that $\frac{(a + 2d)^2}{a + d} - a = 30$ . Multiplying out by the denominator, $(a^2 + 4ad + 4d^2) - a(a + d) = 30(a + d).$ This simplifies to $3ad + 4d^2 = 30a + 30d$ , which upon rearranging yields $2d(2d - 15) = 3a(10 - d)$ .

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 $d > 10$ and $d < \frac{15}{2}$ , which is a contradiction). Then, $d = 8, 9$ . Directly substituting and testing shows that $d \neq 8$ , but that if $d = 9$ then $a = 18$ . Alternatively, note that $3|2d$ or $3|2d-15$ implies that $3|d$ , so only $9$ may work. Hence, the four terms are $18,\ 27,\ 36,\ 48$ , which indeed fits the given conditions. Their sum is $\boxed{129}$ .

Postscript

As another option, $3ad + 4d^2 = 30a + 30d$ could be rewritten as follows:

$d(3a + 4d) = 30(a + d)$

$d(3a + 3d)+ d^2 = 30(a + d)$

$3d(a + d)+ d^2 = 30(a + d)$

$(3d - 30)(a + d)+ d^2 = 0$

$3(d - 10)(a + d)+ d^2 = 0$

This gives another way to prove $d<10$ , and when rewritten one last time:

$3(10 -d)(a + d) = d^2$

shows that $d$ must contain a factor of 3.

-jackshi2006

2003 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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2003 AIME I Problems/Problem 8

Problem 8

Solution

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