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{...") |
m (typo fix) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 3: | Line 3: | ||
<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>. |
| + | |||
== 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}} | ||
Latest 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 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
