Difference between revisions of "2003 AIME II Problems/Problem 8"
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== Solution == | == Solution == | ||
| − | + | If you multiply the corresponding terms of two arithmetic sequences, you get the terms of a quadratic function. Thus, we have a quadratic <math>ax^2+bx+c</math> 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 <math>-72(8)^2+492\cdot8+1020=348</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2003|n=II|num-b=7|num-a=9}} | {{AIME box|year=2003|n=II|num-b=7|num-a=9}} | ||
Revision as of 16:12, 21 January 2008
Problem
Find the eighth term of the sequence 
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 
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 
See also
| 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 | ||
| All AIME Problems and Solutions | ||
