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2007 AMC 12A Problem 7

Problem 7 of 25EasierAlgebra

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

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Solution

Let DD be the common difference. Then a=c−2D,a=c-2D, b=c−D,b=c-D, d=c+D,d=c+D, and e=c+2D,e=c+2D, so a+b+c+d+e=5c.a+b+c+d+e=5c. Thus 5c=30,5c=30, giving c=6.c=6. The other terms cannot be determined: the sequences 4,5,6,7,84,5,6,7,8 and 10,8,6,4,210,8,6,4,2 both satisfy the conditions but differ in every term except the middle one. Thus, the correct answer is C.

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Concepts: arithmetic sequence · symmetry (algebra)

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