Difference between revisions of "2005 AMC 10A Problems/Problem 17"

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Since each number is part of <math>2</math> numbers in the [[arithmetic sequence]], the sum of the arithmetic sequence is <math>2(3+5+6+7+9)=60</math>
 
Since each number is part of <math>2</math> numbers in the [[arithmetic sequence]], the sum of the arithmetic sequence is <math>2(3+5+6+7+9)=60</math>
  
Since the middle term in an arithmetic sequence is the average of the terms, the middle number is <math>\frac{60}{5}=12\Rightarrow D</math>
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Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is <math>\frac{60}{5}=12\Rightarrow D</math>
  
 
==See Also==
 
==See Also==

Revision as of 17:41, 4 August 2006

Problem

In the five-sided star shown, the letters $A$, $B$, $C$, $D$, and $E$ are replaced by the numbers $3$, $5$, $6$, $7$, and $9$, although not necessarily in this order. The sums of the numbers at the ends of the line segments $AB$, $BC$, $CD$, $DE$, and $EA$ form an arithmetic sequence, although not necessarily in this order. What is the middle term of the sequence?

2005amc10a17.gif

$\mathrm{(A) \ } 9\qquad \mathrm{(B) \ } 10\qquad \mathrm{(C) \ } 11\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ } 13$

Solution

Since each number is part of $2$ numbers in the arithmetic sequence, the sum of the arithmetic sequence is $2(3+5+6+7+9)=60$

Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is $\frac{60}{5}=12\Rightarrow D$

See Also