Difference between revisions of "2011 AMC 12A Problems/Problem 8"
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<math>(A-F-G)+(F+G+H)=-5+30 \implies (A+H)=25</math> | <math>(A-F-G)+(F+G+H)=-5+30 \implies (A+H)=25</math> | ||
− | Therefore, we have | + | Therefore, we have <math>A+H=25 \rightarrow \boxed{\textbf{C}}</math> |
~JinhoK | ~JinhoK |
Revision as of 02:38, 12 July 2020
Contents
[hide]Problem
In the eight term sequence , , , , , , , , the value of is and the sum of any three consecutive terms is . What is ?
Solution
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 deduct the following equations:
, and .
We can then cleverly add and subtract the equations above to be left with the answer.
(Notice how we don't use )
Therefore, we have
~JinhoK
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.