A little math.

For a positive integer n, the mean of the first n terms of a sequence is n. What is the 2018th term of this sequence?

4037 4035 2018 4036 4034

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1 solution

The sequence is { a n } n = 1 \{a_n\}_{n=1}^{\infty}

1 2018 i = 1 2018 a i = 2018 \frac{1}{2018}\sum_{i=1}^{2018}a_i=2018

1 2017 i = 1 2017 a i = 2017 \frac{1}{2017}\sum_{i=1}^{2017}a_i=2017

a 2018 = i = 1 2018 a i i = 1 2018 a i = 201 8 2 201 7 2 = 2018 a_{2018}=\sum_{i=1}^{2018}a_i-\sum_{i=1}^{2018}a_i=2018^2-2017^2=2018

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