# 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 $30$. 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

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