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>
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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?
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<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==
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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

$1,-2,3,-4,5,-6,\ldots,$

whose $n$th term is $(-1)^{n+1}\cdot n$. What is the average of the first $200$ terms of the sequence?

$\textbf{(A)}-\!1\qquad\textbf{(B)}-\!0.5\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 0.5\qquad\textbf{(E)}\ 1$

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: $(1 + -2) + (3 + -4) + ... + (199 + -200)$

The sum of each pair is $-1$, and there are $100$ pairs, so the total sum is $-100$.

There are $200$ numbers on the list, so the average is $\frac{-100}{200} = -0.5$, and the answer is $\boxed{B}$

See also

1997 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AHSME Problems and Solutions