1999 AHSME Problems/Problem 1

Problem

$1 - 2 + 3 -4 + \cdots - 98 + 99 =$

$\mathrm{(A) \ -50 } \qquad \mathrm{(B) \ -49 } \qquad \mathrm{(C) \ 0 } \qquad \mathrm{(D) \ 49 } \qquad \mathrm{(E) \ 50 }$

Solution

If we group consecutive terms together, we get $(-1) + (-1) + \cdots + 99$ , and since there are 49 pairs of terms the answer is $-49 + 99 = 50 \Rightarrow \mathrm{(E)}$ .

1999 AHSME (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
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1999 AHSME Problems/Problem 1

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