A sequence consists of 2010 terms. Each term after the first is 1 larger than the previous term. The sum of the 2010 terms is 5307. When every second term is added up, starting with the first term and ending with the second last term, the sum is
(A) 2155 (B) 2153 (C) 2151 (D) 2149 (E) 2147
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puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.
Source: CEMC, University of Waterloo,
licensed CC-BY-NC-4.0.
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