2003 AIME II Problems/Problem 8: Difference between revisions
| Line 39: | Line 39: | ||
Now we find | Now we find | ||
<math>(a+7r_1)(b+7r_2) = ab + 7(ar_2 + br_1) + 49r_1r_2 = 1440 + 7(348) + 49(-72) = \boxed{348}</math> | <math>(a+7r_1)(b+7r_2) = ab + 7(ar_2 + br_1) + 49r_1r_2 = 1440 + 7(348) + 49(-72) = \boxed{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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:29, 17 October 2013
Problem
Find the eighth term of the sequence 
whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.
Solution
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 
such that 
, 
, and 
. Plugging in the values for x gives us a system of three equations:
Solving gives 
and 
. Thus, the answer is 
Solution 2
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 | ||
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