A8. We are looking at a sequence of numbers that starts with $27,1,2012, \ldots$. For this sequence of numbers, the following applies. If we add the numbers in positions 1, 2, and 3 in the sequence, we get 2040. The sum of the numbers in positions 2, 3, and 4 is 2039, and the sum of the numbers in positions 3, 4, and 5 is 2038, and so on. Generally, therefore: the numbers in positions $k, k+1$, and $k+2$ together are $2041-k$. What number is in position 2013 in the sequence?
A) -670
B) -669
C) 670
D) 1341
E) 1342

## B-questions
A8. E) 1342 If $a, b, c, d$ are four consecutive numbers in the sequence, then $d=a-1$, because from the given it follows that $b+c+d=(a+b+c)-1$. In other words: if you move three positions further in the sequence, you find a number that is 1 less than the number where you started. The numbers at positions $3,6,9, \ldots, 2013(=3+3 \times 670)$ are thus exactly $2012,2011, \ldots,(2012-670=) 1342$.