Difference between revisions of "2011 AMC 10A Problems/Problem 4"
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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:  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:  
<math>46\cdot 2=\boxed{92}</math>  <math>46\cdot 2=\boxed{92}</math>  
+  
+  ==Solution 3==  
+  
+  <cmath>\begin{align*}  
+  X&=10+12+14+\cdots +100 \\  
+  Y&=X10+102 = X+92 \\  
+  YX &= (X+92)X \\  
+  YX &= XX+92 \\  
+  YX &= 0+92 \\  
+  YX &= \boxed{92} \quad \quad \textbf{(A)}\\  
+  \end{align*} </cmath>  
+  <math>\blacksquare</math>  
+  
+   <math>\text{herobrineindia}</math>  
+  
== See Also ==  == See Also ==  
{{AMC10 boxyear=2011ab=Anumb=3numa=5}}  {{AMC10 boxyear=2011ab=Anumb=3numa=5}}  
{{MAA Notice}}  {{MAA Notice}} 
Revision as of 02:05, 23 January 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

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.