Difference between revisions of "2011 AMC 10A Problems/Problem 4"
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\end{align*}</cmath> | \end{align*}</cmath> | ||
From here it is obvious that Y - X = 102 - 10 = <math>\boxed{92 \ \mathbf{(A)}}</math>. | From here it is obvious that Y - X = 102 - 10 = <math>\boxed{92 \ \mathbf{(A)}}</math>. | ||
| + | |||
| + | =Note= | ||
| + | Another way to see this is to let the sum <math>12+14+16+...+100=x.</math> So, the sequences become | ||
| + | <cmath>\begin{align*} | ||
| + | X = 10+x \\ | ||
| + | Y= x+102 \\ | ||
| + | \end{align*}</cmath> | ||
| + | |||
| + | Like before, the difference between the two sequences is <math>Y-X=102-12=90.</math> | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 21:25, 30 January 2018
Let X and Y be the following sums of arithmetic sequences:
What is the value of Y - X?
Contents
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 Y - X = 102 - 10 = 
.
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:
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. 
