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2008 AMC 12B Problem 12

Problem 12 of 25IntermediateAlgebraProbability & Statistics

For each positive integer n,n, the mean of the first nn terms of a sequence is n.n. What is the 20082008th term of the sequence?

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Solution

Since the mean of the first nn terms is n,n, their sum is n⋅n=n2.n \cdot n = n^2. The nnth term is the difference of consecutive sums, n2−(n−1)2=2n−1.n^2 - (n-1)^2 = 2n - 1. For n=2008,n = 2008, the term is 2⋅2008−1=4015.2 \cdot 2008 - 1 = 4015. Thus, the correct answer is B.
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