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

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 $-72(8)^2+492\cdot8+1020=348$

2003 AIME II (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
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2003 AIME II Problems/Problem 8

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