Difference between revisions of "2003 AIME II Problems/Problem 8"
Chickendude (talk | contribs) (Added an alternate Solution) |
Giratina150 (talk | contribs) (→Solution) |
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<math>9a+3b+c=1848</math> | <math>9a+3b+c=1848</math> | ||
− | Solving gives a=-72, b=492, and c=1020. Thus, the answer is <math>-72(8)^2+492\cdot8+1020=348</math> | + | Solving gives <math>a=-72, b=492,</math> and <math>c=1020</math>. Thus, the answer is <math>-72(8)^2+492\cdot8+1020=348</math>. |
==Alternate Solution== | ==Alternate Solution== |
Revision as of 00:58, 27 February 2011
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 , , and . Plugging in the values for x gives us a system of three equations:
Solving gives and . Thus, the answer is .
Alternate Solution
Setting one of the sequences as and the other as , we can set up the following equalities
We want to find
Foiling out the two above, we have
and
Plugging in and bringing the constant over yields
Subtracting the two yields and plugging that back in yields
Now we find
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 |