2003 AIME II Problems/Problem 8

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

See also

2003 AIME II (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