A sequence a1,…,a100 of real numbers is such that the arithmetic mean of two consecutive terms is always equal to the index of the second term (for example, we have 2a4+a5=5 ); what is the sum of the 100 numbers of the sequence?
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Solution
Solution:
The answer is (C). Let us denote by S the sum of the 100 terms of the sequence. We have: 2S=21(a1+a2+…+a99+a100)=2a1+a2+2a3+a4+…+2a99+a100. We observe that each of the addends on the rightmost side is the arithmetic mean of two consecutive terms of the sequence, which we know to be equal to the index of the second term. We thus obtain 2S=2+4+…+100=2(1+2+…+50) and recalling that the sum of the first n integers equals 2n(n+1) we conclude that S=4(1+2+ +50)=4 50 51}{2}=5100.
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