Difference between revisions of "1959 AHSME Problems/Problem 18"
Angrybird029 (talk | contribs) (Created page with "== Problem 18== The arithmetic mean (average) of the first <math>n</math> positive integers is: <math>\textbf{(A)}\ \frac{n}{2} \qquad\textbf{(B)}\ \frac{n^2}{2}\qquad\textbf{...") |
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<math>\textbf{(A)}\ \frac{n}{2} \qquad\textbf{(B)}\ \frac{n^2}{2}\qquad\textbf{(C)}\ n\qquad\textbf{(D)}\ \frac{n-1}{2}\qquad\textbf{(E)}\ \frac{n+1}{2} </math> | <math>\textbf{(A)}\ \frac{n}{2} \qquad\textbf{(B)}\ \frac{n^2}{2}\qquad\textbf{(C)}\ n\qquad\textbf{(D)}\ \frac{n-1}{2}\qquad\textbf{(E)}\ \frac{n+1}{2} </math> | ||
== Solution == | == Solution == | ||
| − | The sum of the first <math>n</math> positive integers is the same as the nth triangular number, which can be expressed as <math> \frac{(n)(n+1)}{2} </math>. Since the question is asking for the arithmetic mean, the whole sum is divided by <math>n</math>, thus giving us <math> \frac {\frac{(n)(n+1)}{2}}{n} </math>, which then simplifies to <math>\textbf{(E)} \frac{n+1}{2}</math> | + | The sum of the first <math>n</math> positive integers is the same as the nth triangular number, which can be expressed as <math> \frac{(n)(n+1)}{2} </math>. Since the question is asking for the arithmetic mean, the whole sum is divided by <math>n</math>, thus giving us <math> \frac {\frac{(n)(n+1)}{2}}{n} </math>, which then simplifies to <math>\boxed{\textbf{(E) } \frac{n+1}{2}}</math> |
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== See also == | == See also == | ||
{{AHSME 50p box|year=1959|num-b=17|num-a=19}} | {{AHSME 50p box|year=1959|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:30, 21 July 2024
Problem 18
The arithmetic mean (average) of the first 
positive integers is:
Solution
The sum of the first positive integers is the same as the nth triangular number, which can be expressed as 
. Since the question is asking for the arithmetic mean, the whole sum is divided by , thus giving us 
, which then simplifies to 
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
