1969 AHSME Problems/Problem 9

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Problem

The arithmetic mean (ordinary average) of the fifty-two successive positive integers beginning at 2 is:

$\text{(A) } 27\quad \text{(B) } 27\tfrac{1}{4}\quad \text{(C) } 27\tfrac{1}{2}\quad \text{(D) } 28\quad \text{(E) } 27\tfrac{1}{2}$

Solution

To solve the problem, find the sum of the first $52$ terms of an arithmetic sequence with first term $2$ and common difference $1$ and divide that by $52$. The $52^\text{nd}$ term of the sequence is $2+51=53$, so the sum of the first $52$ terms of the sequence is $\frac{52(2+53)}{2} = 1430$. Thus, the arithmetic mean is $\frac{1430}{52} = \frac{55}{2} = \boxed{\textbf{(C) } 27\frac{1}{2}}$.

See also

1969 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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
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