Difference between revisions of "1962 AHSME Problems/Problem 8"

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==Problem==
 
==Problem==
  
Given the set of n<math></math> numbers; <math>n > 1</math>, of which one is <math>1 - \frac {1}{n}</math> and all the others are <math>1</math>. The arithmetic mean of the <math>n</math> numbers is:  
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Given the set of <math>n</math> numbers; <math>n > 1</math>, of which one is <math>1 - \frac {1}{n}</math> and all the others are <math>1</math>. The arithmetic mean of the <math>n</math> numbers is:  
  
 
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ n-\frac{1}{n}\qquad\textbf{(C)}\ n-\frac{1}{n^2}\qquad\textbf{(D)}\ 1-\frac{1}{n^2}\qquad\textbf{(E)}\ 1-\frac{1}{n}-\frac{1}{n^2} </math>
 
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ n-\frac{1}{n}\qquad\textbf{(C)}\ n-\frac{1}{n^2}\qquad\textbf{(D)}\ 1-\frac{1}{n^2}\qquad\textbf{(E)}\ 1-\frac{1}{n}-\frac{1}{n^2} </math>

Revision as of 22:31, 9 November 2013

Problem

Given the set of $n$ numbers; $n > 1$, of which one is $1 - \frac {1}{n}$ and all the others are $1$. The arithmetic mean of the $n$ numbers is:

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ n-\frac{1}{n}\qquad\textbf{(C)}\ n-\frac{1}{n^2}\qquad\textbf{(D)}\ 1-\frac{1}{n^2}\qquad\textbf{(E)}\ 1-\frac{1}{n}-\frac{1}{n^2}$

Solution

"Unsolved"