AoPSWiki
Do you have what it takes to be the next brilliant trader, researcher, or developer at Jane Street Capital? Find out in the Careers in Mathematics Forum.
Personal tools

2007 AMC 12A Problems/Problem 7

From AoPSWiki

Problem

Let a, b, c, d, and e be five consecutive terms in an arithmetic sequence, and suppose that a+b+c+d+e=30. Which of a, b, c, d, or e can be found?

\textrm{(A)} \ a\qquad \textrm{(B)}\ b\qquad \textrm{(C)}\ c\qquad \textrm{(D)}\ d\qquad \textrm{(E)}\ e

Solution

Let f be the common difference between the terms.

  • a=c-2f
  • b=c-f
  • c=c
  • d=c+f
  • e=c+2f

a+b+c+d+e=5c=30, so c=6. But we can't find any more variables, because we don't know what f is. So the answer is \textrm{C}.

See also

2007 AMC 12A (ProblemsResources)
Preceded by
Problem 6
Followed by
Problem 8
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
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us