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> | ||
+ | |||
+ | == Solution 5 (assumption) == | ||
+ | |||
+ | Assume the sequence is <cmath>15,10,5,15,10,5,15,10</cmath>. | ||
+ | |||
+ | Thus, <math>15+10=\boxed{25}</math> or option <math>\boxed{\mathbf{(C )}25}</math> | ||
+ | |||
+ | ~SirAppel | ||
+ | |||
+ | == Solution 6 == | ||
+ | |||
+ | Notice that the period of the sequence is <math>3</math> as given. (If this isn't clear we can show an example: <math>A+B+C=B+C=D</math> <math>\Leftrightarrow</math> <math>A=D</math>). Then <math>A=D</math> and <math>H=B</math>, so <math>A+H=D+B=D+B+C-C=30-5=\boxed{\textbf{(C)}\;25}</math>. | ||
+ | |||
+ | ~eevee9406 | ||
+ | |||
+ | ==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 | ||
==Video Solution== | ==Video Solution== | ||
Line 50: | Line 67: | ||
https://www.youtube.com/watch?v=6tlqpAcmbz4 | https://www.youtube.com/watch?v=6tlqpAcmbz4 | ||
~Shreyas S | ~Shreyas S | ||
+ | |||
+ | ==Podcast Solution== | ||
+ | |||
+ | https://www.buzzsprout.com/56982/episodes/383730 (Episode starts with a solution to this question) | ||
+ | —[[User:wescarroll|wescarroll]] | ||
== See also == | == See also == |
Latest revision as of 23:12, 8 April 2024
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
Solution 5 (assumption)
Assume the sequence is .
Thus, or option
~SirAppel
Solution 6
Notice that the period of the sequence is as given. (If this isn't clear we can show an example:
). Then
and
, so
.
~eevee9406
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
Podcast Solution
https://www.buzzsprout.com/56982/episodes/383730 (Episode starts with a solution to this question) —wescarroll
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.