Difference between revisions of "2011 AMC 12A Problems/Problem 8"
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Since all of the answer choices are constants, it shouldn't matter what we pick <math>A</math> and <math>B</math> to be, so let <math>A = 20</math> and <math>B = 5</math>. Then <math>D = 30 - B -C = 20</math>, <math>E = 30 - D - C = 5</math>, <math>F = 30 - D - E =5</math>, and so on until we get <math>H = 5</math>. Thus <math>A + H = \boxed{\mathbf{(C)}25}</math> | Since all of the answer choices are constants, it shouldn't matter what we pick <math>A</math> and <math>B</math> to be, so let <math>A = 20</math> and <math>B = 5</math>. Then <math>D = 30 - B -C = 20</math>, <math>E = 30 - D - C = 5</math>, <math>F = 30 - D - E =5</math>, and so on until we get <math>H = 5</math>. Thus <math>A + H = \boxed{\mathbf{(C)}25}</math> | ||
− | =Note= | + | ==Note== |
Something useful to shorten a lot of the solutions above is to notice <cmath>5 + D + E = D + E + F</cmath> so F = 5 | Something useful to shorten a lot of the solutions above is to notice <cmath>5 + D + E = D + E + F</cmath> so F = 5 | ||
Revision as of 21:12, 31 May 2022
Contents
Problem
In the eight term sequence , , , , , , , , the value of is and the sum of any three consecutive terms is . What is ?
Solution 1
Let . Then from , we find that . From , we then get that . Continuing this pattern, we find , , , and finally . So
Solution 2
Given that the sum of 3 consecutive terms is 30, we have and
It follows that because .
Subtracting, we have that .
Solution 3 (the tedious one)
From the given information, we can deduce the following equations:
, and .
We can then cleverly manipulate the equations above by adding and subtracting them to be left with the answer.
(Notice how we don't use )
Therefore, we have
~JinhoK
Solution 4 (the cheap one)
Since all of the answer choices are constants, it shouldn't matter what we pick and to be, so let and . Then , , , and so on until we get . Thus
Note
Something useful to shorten a lot of the solutions above is to notice so F = 5
Video Solution
https://www.youtube.com/watch?v=6tlqpAcmbz4 ~Shreyas S
See also
2011 AMC 12A (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.