Difference between revisions of "2011 AMC 10A Problems/Problem 17"
(→Solution 2) |
(→Solution) |
||
Line 14: | Line 14: | ||
Equating the two values we get for the sum, we get the answer <math>A+H+60=85</math> <math>\Longrightarrow</math> <math>A+H=\boxed{25 \ \mathbf{(C)}}</math>. | Equating the two values we get for the sum, we get the answer <math>A+H+60=85</math> <math>\Longrightarrow</math> <math>A+H=\boxed{25 \ \mathbf{(C)}}</math>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Let <math>A=x</math>. Then from <math>A+B+C=30</math>, we find that <math>B=25-x</math>. From <math>B+C+D=30</math>, we then get that <math>D=x</math>. Continuing this pattern, we find <math>E=25-x</math>, <math>F=5</math>, <math>G=x</math>, and finally <math>H=25-x</math>. So <math>A+H=x+25-x=25 \rightarrow \boxed{\textbf{C}}</math> | ||
+ | |||
+ | ===Solution 4=== | ||
+ | Given that the sum of 3 consecutive terms is 30, we have | ||
+ | <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 <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>. | ||
== Solution 2 == | == Solution 2 == |
Revision as of 02:59, 11 January 2018
Problem 17
In the eight-term sequence , the value of
is 5 and the sum of any three consecutive terms is 30. What is
?
Solution
We consider the sum and use the fact that
, and hence
.
Equating the two values we get for the sum, we get the answer
.
Solution 3
Let . Then from
, we find that
. From
, we then get that
. Continuing this pattern, we find
,
,
, and finally
. So
Solution 4
Given that the sum of 3 consecutive terms is 30, we have
and
It follows that because
.
Subtracting, we have that .
Solution 2
We see that , and by substituting the given
, we find that
. Similarly,
and
.
Similarly, and
, giving us
. Since
,
.
See Also
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.