A1. Julia creates a sequence of numbers. She starts with two self-chosen numbers. Then she calculates the next number in the sequence as follows: if the last number she wrote down is $b$ and the number before that is $a$, then the next number is $2b - a$. The second number in Julia's sequence is 55 and the hundredth number is 2015.

What is the first number in her sequence?
B1. 35

When $a, b, c$ are three consecutive numbers from Julia's sequence, we know that $c=2 b-a$. In other words: $c-b=b-a$. This means that the difference between two consecutive numbers in the sequence is always the same, say $d$. If the first number in the sequence is $x$, then the sequence is equal to $x, x+d, x+2 d, x+3 d, \ldots$. The second number in the sequence is equal to $55=x+d$ and the hundredth number is $2015=x+99 d$. We see that $2015-55=98 d$, so $d=\frac{1960}{98}=20$. With this, we find that $55=x+20$, so $x=35$. The first number in Julia's sequence is therefore 35.