# Difference between revisions of "2003 AIME II Problems/Problem 8"

## Problem

Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.

## Solution 1

If you multiply the corresponding terms of two arithmetic sequences, you get the terms of a quadratic function. Thus, we have a quadratic $ax^2+bx+c$ such that $f(1)=1440$, $f(2)=1716$, and $f(3)=1848$. Plugging in the values for x gives us a system of three equations: $a+b+c=1440$ $4a+2b+c=1716$ $9a+3b+c=1848$

Solving gives $a=-72, b=492,$ and $c=1020$. Thus, the answer is $-72(8)^2+492\cdot8+1020= \boxed{348}.$

## Solution 2

Setting one of the sequences as $a+nr_1$ and the other as $b+nr_2$, we can set up the following equalities $ab = 1440$ $(a+r_1)(b+r_2)=1716$ $(a+2r_1)(b+2r_2)=1848$

We want to find $(a+7r_1)(b+7r_2)$

Foiling out the two above, we have $ab + ar_2 + br_1 + r_1r_2 = 1716$ and $ab + 2ar_2 + 2br_1 + 4r_1r_2 = 1848$

Plugging in $ab=1440$ and bringing the constant over yields $ar_2 + br_1 + r_1r_2 = 276$ $ar_2 + br_1 + 2r_1r_2 = 204$

Subtracting the two yields $r_1r_2 = -72$ and plugging that back in yields $ar_2 + br_1 = 348$

Now we find $(a+7r_1)(b+7r_2) = ab + 7(ar_2 + br_1) + 49r_1r_2 = 1440 + 7(348) + 49(-72) = \boxed{348}$.

## Solution 3

Let the first sequence be $A={a+d_1, a + 2d_1, a + 3d_1, \cdots}$

and the second be $B={b+d_2, b + 2d_2, b + 3d_2, \cdots}$,

with $(a+d_1)(b+d_2)=1440$. Now, note that the $n^{\text{th}}$ term of sequence $A$ is $a+d_1 n$ and the $n^{\text{th}}$ term of $B$ is $b + d_2 n$. Thus, the $n^{\text{th}}$ term of the given sequence is $n^2(d_1 + d_2) + n(ad_2 + bd_1) + ab$,

a quadratic in $n$. Now, letting the given sequence be $C$, we see that $C_n - C_{n-1} = n^2(d_1 + d_2) + n(ad_2 + bd_1) + ab - (n-1)^2(d_1 + d_2) - (n-1)(ad_2 + bd_1) - ab = n(2d_1 + 2d_2) + ad_2 + bd_1 - d_1 - d_2$,

a linear equation in $n$! Since $C_2 - C_1 = 276$ and $C_3 - C_2 = 132$, we can see that, in general, we have $C_n - C_{n-1} = 420 - 144n$.

Thus, we can easily find $C_4 - C_3 = -12 \rightarrow C_4 = 1836$, $C_5 - C_4 = -156 \rightarrow C_5 = 1680$, $C_6 - C_5 = -300 \rightarrow C_6 = 1380$, $C_7 - C_6 = -444 \rightarrow C_7 = 936$, and finally $C_8 - C_7 = -588 \rightarrow \boxed{C_8 = 348}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 