1959 AHSME Problems/Problem 18

Problem 18

The arithmetic mean (average) of the first $n$ positive integers is: $\textbf{(A)}\ \frac{n}{2} \qquad\textbf{(B)}\ \frac{n^2}{2}\qquad\textbf{(C)}\ n\qquad\textbf{(D)}\ \frac{n-1}{2}\qquad\textbf{(E)}\ \frac{n+1}{2}$

Solution

The sum of the first $n$ positive integers is the same as the nth triangular number, which can be expressed as $\frac{(n)(n+1)}{2}$ . Since the question is asking for the arithmetic mean, the whole sum is divided by $n$ , thus giving us $\frac {\frac{(n)(n+1)}{2}}{n}$ , which then simplifies to $\boxed{\textbf{(E) } \frac{n+1}{2}}$

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1959 AHSME Problems/Problem 18

Problem 18

Solution

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