2011 AMC 12A Problems/Problem 8
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 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination