Difference between revisions of "2011 AMC 12A Problems/Problem 8"
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It follows that, because <math>C=5</math>, <math>A+B+C+D+E+F+G+H=85</math>. | It follows that, because <math>C=5</math>, <math>A+B+C+D+E+F+G+H=85</math>. | ||
− | Subtracting, we have that <math>A+H=25</math>. | + | Subtracting, we have that <math>A+H=25\rightarrow \boxed{\textbf{C}}</math>. |
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=7|num-a=9|ab=A}} | {{AMC12 box|year=2011|num-b=7|num-a=9|ab=A}} |
Revision as of 07:11, 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 |