Difference between revisions of "2011 AMC 10A Problems/Problem 17"
(→Solution 2) |
|||
| Line 36: | Line 36: | ||
Similarly, <math>G=A</math> and <math>H=B</math>, giving us <math>A, B, 5, A, B, 5, A, B</math>. Since <math>H=B</math>, <math>A+H=A+B=\boxed{25 \ \mathbf{(C)}}</math>. | Similarly, <math>G=A</math> and <math>H=B</math>, giving us <math>A, B, 5, A, B, 5, A, B</math>. Since <math>H=B</math>, <math>A+H=A+B=\boxed{25 \ \mathbf{(C)}}</math>. | ||
| + | |||
| + | == Solution 4 == | ||
| + | Arbitrarily select a value for B or D, and then deduce the rest using the fact that consecutive triplets add to 30. You will find that <math>A+H= \boxed{\textbf{25}}</math>. | ||
== See Also == | == See Also == | ||
Revision as of 13:51, 26 January 2020
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 2
Given that the sum of 3 consecutive terms is 30, we have
and 
It follows that 
because .
Subtracting, we have that 
.
Solution 3
We see that 
, and by substituting the given , we find that . Similarly, 
and 
.
Similarly, 
and 
, giving us 
. Since , 
.
Solution 4
Arbitrarily select a value for B or D, and then deduce the rest using the fact that consecutive triplets add to 30. You will find that 
.
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. 
