Difference between revisions of "1969 AHSME Problems/Problem 9"
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\text{(C) } 27\tfrac{1}{2}\quad | \text{(C) } 27\tfrac{1}{2}\quad | ||
\text{(D) } 28\quad | \text{(D) } 28\quad | ||
| − | \text{(E) } | + | \text{(E) } 28\tfrac{1}{2}</math> |
== Solution == | == Solution == | ||
| − | <math>\ | + | To solve the problem, find the sum of the first <math>52</math> terms of an [[arithmetic sequence]] with first term <math>2</math> and common difference <math>1</math> and divide that by <math>52</math>. The <math>52^\text{nd}</math> term of the sequence is <math>2+51=53</math>, so the sum of the first <math>52</math> terms of the sequence is <math>\frac{52(2+53)}{2} = 1430</math>. Thus, the [[arithmetic mean]] is <math>\frac{1430}{52} = \frac{55}{2} = \boxed{\textbf{(C) } 27\frac{1}{2}}</math>. |
== See also == | == See also == | ||
Latest revision as of 14:42, 13 November 2024
Problem
The arithmetic mean (ordinary average) of the fifty-two successive positive integers beginning at 2 is:
Solution
To solve the problem, find the sum of the first 
terms of an arithmetic sequence with first term 
and common difference 
and divide that by . The 
term of the sequence is 
, so the sum of the first terms of the sequence is 
. Thus, the arithmetic mean is 
.
See also
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
