Difference between revisions of "1997 AHSME Problems/Problem 6"
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+ | ==Problem== | ||
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+ | Consider the sequence | ||
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+ | <math> 1,-2,3,-4,5,-6,\ldots, </math> | ||
+ | |||
+ | whose <math>n</math>th term is <math> (-1)^{n+1}\cdot n </math>. What is the average of the first <math>200</math> terms of the sequence? | ||
+ | |||
+ | <math> \textbf{(A)}-\!1\qquad\textbf{(B)}-\!0.5\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 0.5\qquad\textbf{(E)}\ 1 </math> | ||
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+ | ==Solution== | ||
+ | |||
+ | The average of a list is the sum of all numbers divided by the size of the list. | ||
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+ | The sum of the list can be found by adding the numbers in pairs: <math>(1 + -2) + (3 + -4) + ... + (199 + -200)</math> | ||
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+ | The sum of each pair is <math>-1</math>, and there are <math>100</math> pairs, so the total sum is <math>-100</math>. | ||
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+ | There are <math>200</math> numbers on the list, so the average is <math>\frac{-100}{200} = -0.5</math>, and the answer is <math>\boxed{B}</math> | ||
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== See also == | == See also == | ||
{{AHSME box|year=1997|num-b=5|num-a=7}} | {{AHSME box|year=1997|num-b=5|num-a=7}} |
Revision as of 16:54, 8 August 2011
Problem
Consider the sequence
whose th term is . What is the average of the first terms of the sequence?
Solution
The average of a list is the sum of all numbers divided by the size of the list.
The sum of the list can be found by adding the numbers in pairs:
The sum of each pair is , and there are pairs, so the total sum is .
There are numbers on the list, so the average is , and the answer is
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |