Difference between revisions of "2011 AMC 10A Problems/Problem 4"
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- <math>\text{herobrine-india}</math> | - <math>\text{herobrine-india}</math> | ||
| + | ==Video Solution== | ||
| + | https://youtu.be/L6utIF9FzPQ | ||
| + | |||
| + | ~savannahsolver | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2011|ab=A|num-b=3|num-a=5}} | {{AMC10 box|year=2011|ab=A|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:02, 24 October 2020
Problem
Let X and Y be the following sums of arithmetic sequences:
What is the value of 
Solution 1
We see that both sequences have equal numbers of terms, so reformat the sequence to look like:
From here it is obvious that 
.
Note
Another way to see this is to let the sum 
So, the sequences become
Like before, the difference between the two sequences is 
Solution 2
We see that every number in Y's sequence is two more than every corresponding number in X's sequence. Since there are 46 numbers in each sequence, the difference must be:
Solution 3
- 
Video Solution
~savannahsolver
See Also
| 2011 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
