Difference between revisions of "2011 AMC 10A Problems/Problem 4"
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m (→Solution 3) |
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Y&=X-10+102 = X+92 \\ | Y&=X-10+102 = X+92 \\ | ||
Y-X &= (X+92)-X \\ | Y-X &= (X+92)-X \\ | ||
| − | + | &= \boxed{92} \quad \quad \textbf{(A)}\\ | |
| − | |||
| − | |||
\end{align*} </cmath> | \end{align*} </cmath> | ||
<math>\blacksquare</math> | <math>\blacksquare</math> | ||
Revision as of 11:05, 19 October 2023
Contents
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
- 
Solution 4
In an actual contest, this would probably take too much time but is nevertheless a solution.
The general formula for computing sums of any arithmetic sequence where 
is the number of terms, 
is the first term and 
is the last term is 
. If one uses that formula for both sequences, they will get 
as the sum for 
and 
as the sum for 
.
Subtracting from , one will get the answer 
.
- danfan
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. 
