2007 AMC 12A Problem 7

Attempt Problem 7 of the 2007 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12A solutions, or check the answer key.

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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 the following can be found?

$a$

$b$

$c$

$d$

$e$

Answer: C

Concepts:arithmetic sequence symmetry (algebra)

Difficulty rating: 1130

Solution:

Let $D$ be the common difference. Then $a=c-2D,$ $b=c-D,$ $d=c+D,$ and $e=c+2D,$ so $a+b+c+d+e=5c.$

Thus $5c=30,$ giving $c=6.$

The other terms cannot be determined: the sequences $4,5,6,7,8$ and $10,8,6,4,2$ both satisfy the conditions but differ in every term except the middle one.

Thus, the correct answer is C.

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