Difference between revisions of "2011 AMC 12A Problems/Problem 8"
(→Solution 3 (the tedious one)) |
|||
Line 23: | Line 23: | ||
==Solution 3 (the tedious one)== | ==Solution 3 (the tedious one)== | ||
− | From the given information, we can | + | From the given information, we can deduce the following equations: |
<math>A+B=25, B+D=25, D+E=25, D+E+F=30, E+F+G | <math>A+B=25, B+D=25, D+E=25, D+E+F=30, E+F+G |
Latest revision as of 14:57, 12 July 2021
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
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.