Difference between revisions of "2003 AIME II Problems/Problem 8"
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Sequence 2:<math>b,b+d_2,b+2d_2,\dots b+(n-1)d_2</math>. | Sequence 2:<math>b,b+d_2,b+2d_2,\dots b+(n-1)d_2</math>. | ||
− | Additionally, | + | Additionally, label the sequence given in the problem the function <math>f</math>, such that |
<math>f(1)=1440,f(2)=1716,f(3)=1848</math>. | <math>f(1)=1440,f(2)=1716,f(3)=1848</math>. | ||
− | Then, <math>f(1)=ab,</math> <math>f(2)=(a+d_1)(b+d_2),</math> | + | Then, |
+ | |||
+ | <math>f(1)=ab,</math> | ||
+ | |||
+ | <math>f(2)=(a+d_1)(b+d_2)=a+ad_2+d_1b+d_1d_2,</math> | ||
+ | |||
+ | <math>f(3)=(a+2d_1)(b+2d_2)=ab+2ad_2+2bd_1+4d_1d_2,</math> | ||
+ | |||
+ | and <math>f(8)=(a+7d_1)(b+7d_2)=ab+7ad_2+7bd_1+49d_1d_2</math>. | ||
+ | |||
+ | In order to find <math>f(8)</math> add <math>f(3)</math> enough times to get the difference between the <math>d_1d_2</math> and <math>ad_2+bd_1</math> terms, then add <math>f(2)</math> and <math>f(1)</math> to get the other terms: | ||
+ | |||
+ | <math>21f(3)=21ab+42ad_2+42bd_1+84d_1d_2</math> | ||
+ | |||
+ | <math>21f(3)-35f(2)=21ab+42ad_2+42bd_1+84d_1d_2-35a-35ad_2-35d_1b-35d_1d_2=-14ab+7ad_2+7bd_1+49d_1d_2</math> | ||
+ | |||
+ | <math>21f(3)-35f(2)+15f(1)=-14ab+7ad_2+7bd_1+49d_1d_2+15ab=ab+7ad_2+7bd_1+49d_1d_2</math> | ||
+ | |||
+ | Now that the expression is in terms of the given values, insert values and solve: | ||
+ | |||
+ | <math>21*1848-35*1716+15*1440=1848+20*132+(20*1716-35*1716+15*1716)+15*(-276)</math> | ||
+ | |||
+ | <math>=1848+5*132+15(132-276)</math> | ||
+ | |||
+ | <math>=1848+660+15(-144)</math> | ||
+ | |||
+ | <math>=348</math> | ||
==See also == | ==See also == |
Latest revision as of 14:31, 24 June 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
.
Solution 4(Tedious)
Start by labeling the two sequences:
Sequence 1:,
Sequence 2:.
Additionally, label the sequence given in the problem the function , such that
.
Then,
and .
In order to find add enough times to get the difference between the and terms, then add and to get the other terms:
Now that the expression is in terms of the given values, insert values and solve:
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.