Difference between revisions of "1997 AHSME Problems/Problem 6"

Latest revision as of 14:12, 5 July 2013

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?

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

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 $\boxed{B}$

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

@@ Line 1: / Line 1: @@
+==Problem==
+Consider the sequence
+<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>
+==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:  <math>(1 + -2) + (3 + -4) + ... + (199 + -200)</math>
+The sum of each pair is <math>-1</math>, and there are <math>100</math> pairs, so the total sum is <math>-100</math>.
+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>
 == See also ==
 {{AHSME box|year=1997|num-b=5|num-a=7}}
+{{MAA Notice}}