Difference between revisions of "2011 AMC 12A Problems/Problem 8"
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<math>(A+B+C)+(C+D+E)+(F+G+H)=90</math> and <math>(B+C+D)+(E+F+G)=60</math> | <math>(A+B+C)+(C+D+E)+(F+G+H)=90</math> and <math>(B+C+D)+(E+F+G)=60</math> | ||
| − | It follows that | + | It follows that <math>A+B+C+D+E+F+G+H=85</math> because <math>C=5</math>. |
Subtracting, we have that <math>A+H=25\rightarrow \boxed{\textbf{C}}</math>. | Subtracting, we have that <math>A+H=25\rightarrow \boxed{\textbf{C}}</math>. | ||
Revision as of 06:12, 7 February 2012
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
A faster technique is to assume that the problem can be solved, and thus is an invariant. Since 
, assign any value to . 
is a simple value to plug in, which gives a value of 
for B. The 8-term sequence is thus 
. The sum of the first and the last terms is 
Note that this alternate solution is not a proof. If the sum of 
had been asked for, this technique would have given 
as an answer, when the true answer would have been "cannot be determined".
Solution 3
Given that the sum of 3 consecutive terms is 30, we have
and 
It follows that 
because 
.
Subtracting, we have that 
.
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 | |
