1969 AHSME Problems/Problem 9

Problem

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

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

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 $\frac{1430}{52} = \frac{55}{2} = \boxed{\textbf{(C) } 27\frac{1}{2}}$ .

1969 AHSC (Problems • Answer Key • Resources)
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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1969 AHSME Problems/Problem 9

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