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

(Video Solution)
(Solution)
Line 10: Line 10:
  
 
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}=12\Rightarrow D</math>
 +
 +
 +
== Video Solution ==
 +
https://youtu.be/tKsYSBdeVuw?t=544
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Revision as of 22:41, 27 January 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

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

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 $\frac{60}{5}=12\Rightarrow D$


Video Solution

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

~ pi_is_3.14

See Also

2005 AMC 10A (ProblemsAnswer KeyResources)
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