Difference between revisions of "2003 AIME II Problems/Problem 8"
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− | == Problem == | + | ==Problem== |
Find the eighth term of the sequence <math>1440,</math> <math>1716,</math> <math>1848,\ldots,</math> whose terms are formed by multiplying the corresponding terms of two arithmetic sequences. | Find the eighth term of the sequence <math>1440,</math> <math>1716,</math> <math>1848,\ldots,</math> whose terms are formed by multiplying the corresponding terms of two arithmetic sequences. | ||
− | == Solution 1 == | + | ==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 <math>ax^2+bx+c</math> such that <math>f(1)=1440</math>, <math>f(2)=1716</math>, and <math>f(3)=1848</math>. Plugging in the values for x gives us a system of three equations: | 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 <math>f(1)=1440</math>, <math>f(2)=1716</math>, and <math>f(3)=1848</math>. Plugging in the values for x gives us a system of three equations: | ||
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<math>C_8 - C_7 = -588 \rightarrow \boxed{C_8 = 348}</math>. | <math>C_8 - C_7 = -588 \rightarrow \boxed{C_8 = 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}} | ||
[[Category: Intermediate Algebra Problems]] | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:53, 18 April 2021
Problem
Find the eighth term of the sequence whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.
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
.
Solution 3
Let the first sequence be
and the second be
,
with . Now, note that the term of sequence is and the term of is . Thus, the term of the given sequence is
,
a quadratic in . Now, letting the given sequence be , we see that
,
a linear equation in ! Since and , we can see that, in general, we have
.
Thus, we can easily find
,
,
,
, and finally
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.