Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 9: Line 9:
 
Each corner <math>(A,B,C,D,E)</math> goes to two sides/numbers. (<math>A</math> goes to <math>AE</math> and <math>AB</math>, <math>D</math> goes to <math>DC</math> and <math>DE</math>). The sum of every term is equal to <math>2(3+5+6+7+9)=60</math>
 
Each corner <math>(A,B,C,D,E)</math> goes to two sides/numbers. (<math>A</math> goes to <math>AE</math> and <math>AB</math>, <math>D</math> goes to <math>DC</math> and <math>DE</math>). The sum of every term is equal to <math>2(3+5+6+7+9)=60</math>
  
−
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>
+
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}=\boxed{\textbf{(D) }12}</math>
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 11:39, 14 December 2021

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 that order. What is the middle term of the arithmetic sequence?

2005amc10a17.gif

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

Solution

Each corner $(A,B,C,D,E)$ goes to two sides/numbers. ($A$ goes to $AE$ and $AB$, $D$ goes to $DC$ and $DE$). The sum of every term is equal to $2(3+5+6+7+9)=60$

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

Video Solution

https://youtu.be/tKsYSBdeVuw?t=544

~ pi_is_3.14

See Also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_17&oldid=167681"