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== | |
− | |||
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: | ||
Line 12: | Line 11: | ||
<math>9a+3b+c=1848</math> | <math>9a+3b+c=1848</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> | + | Solving gives <math>a=-72, b=492,</math> and <math>c=1020</math>. Thus, the answer is <math>-72(8)^2+492\cdot8+1020= \boxed{348}.</math> |
− | === Solution 2 | + | ==Solution 1 (faster)== |
+ | |||
+ | Use the same rationale as in solution 1; instead of using terms <math>1,2,3</math>, we use <math>-1,0,1</math> and solve the <math>6</math>th term. | ||
+ | |||
+ | <math>a-b+c=1440</math> | ||
+ | |||
+ | <math>c=1716</math> | ||
+ | |||
+ | <math>a+b+c=1848</math> | ||
+ | |||
+ | Accordingly we will solve | ||
+ | |||
+ | <math>a=-72, b=204, c=1716</math> | ||
+ | |||
+ | <math>36a+6b+c= \boxed{348}.</math> | ||
+ | |||
+ | -maxamc | ||
+ | |||
+ | == Solution 2 == | ||
Setting one of the sequences as <math>a+nr_1</math> and the other as <math>b+nr_2</math>, we can set up the following equalities | Setting one of the sequences as <math>a+nr_1</math> and the other as <math>b+nr_2</math>, we can set up the following equalities | ||
Line 39: | Line 56: | ||
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>. |
+ | |||
+ | ==Solution 3== | ||
+ | Let the first sequence be | ||
+ | |||
+ | <math>A={a+d_1, a + 2d_1, a + 3d_1, \cdots}</math> | ||
+ | |||
+ | and the second be | ||
+ | |||
+ | <math>B={b+d_2, b + 2d_2, b + 3d_2, \cdots}</math>, | ||
+ | |||
+ | with <math>(a+d_1)(b+d_2)=1440</math>. Now, note that the <math>n^{\text{th}}</math> term of sequence <math>A</math> is <math>a+d_1 n</math> and the <math>n^{\text{th}}</math> term of <math>B</math> is <math>b + d_2 | ||
+ | n</math>. Thus, the <math>n^{\text{th}}</math> term of the given sequence is | ||
+ | |||
+ | <math>n^2(d_1 + d_2) + n(ad_2 + bd_1) + ab</math>, | ||
+ | |||
+ | a quadratic in <math>n</math>. Now, letting the given sequence be <math>C</math>, we see that | ||
+ | |||
+ | <math>C_n - C_{n-1} = n^2(d_1 + d_2) + n(ad_2 + bd_1) + ab - (n-1)^2(d_1 + d_2) - (n-1)(ad_2 + bd_1) - ab = n(2d_1 + 2d_2) + ad_2 + bd_1 - d_1 - d_2</math>, | ||
+ | |||
+ | a linear equation in <math>n</math>. Since <math>C_2 - C_1 = 276</math> and <math>C_3 - C_2 = 132</math>, we can see that, in general, we have | ||
+ | |||
+ | <math>C_n - C_{n-1} = 420 - 144n</math>. | ||
+ | |||
+ | Thus, we can easily find | ||
+ | |||
+ | <math>C_4 - C_3 = -12 \rightarrow C_4 = 1836</math>, | ||
+ | |||
+ | <math>C_5 - C_4 = -156 \rightarrow C_5 = 1680</math>, | ||
+ | |||
+ | <math>C_6 - C_5 = -300 \rightarrow C_6 = 1380</math>, | ||
+ | |||
+ | <math>C_7 - C_6 = -444 \rightarrow C_7 = 936</math>, and finally | ||
+ | |||
+ | <math>C_8 - C_7 = -588 \rightarrow \boxed{C_8 = 348}</math>. | ||
+ | |||
+ | ==Solution 4(Tedious) == | ||
+ | Start by labeling the two sequences: | ||
+ | |||
+ | Sequence 1:<math>a,a+d_1,a+2d_1,\dots a+(n-1)d_1</math>, | ||
+ | |||
+ | Sequence 2:<math>b,b+d_2,b+2d_2,\dots b+(n-1)d_2</math>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | Then, | ||
+ | |||
+ | <math>f(1)=ab,</math> | ||
+ | |||
+ | <math>f(2)=(a+d_1)(b+d_2)=a+ad_2+d_1b+d_1d_2,</math> | ||
− | == See also == | + | <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> | ||
+ | |||
+ | ==Video Solution by Sal Khan== | ||
+ | https://www.youtube.com/watch?v=ZFN63oTeYzc&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=12 - AMBRIGGS | ||
+ | |||
+ | ==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]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:55, 10 July 2024
Contents
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 1 (faster)
Use the same rationale as in solution 1; instead of using terms , we use
and solve the
th term.
Accordingly we will solve
-maxamc
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:
Video Solution by Sal Khan
https://www.youtube.com/watch?v=ZFN63oTeYzc&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=12 - AMBRIGGS
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.