999 numbers are written between 1 and 2018 so that the 1001 numbers form an arithmetic sequence. Find the 501st term.

Step-by-step solution

1. 1

   Fix the index of the last term. With $999$ insertions the sequence runs $u_1=1$ up to $u_{1001}=2018$, so there are $1001-1=1000$ gaps.

2. 2

   Solve for the common difference. $2018=1+1000d\quad\Longrightarrow\quad d=\frac{2017}{1000}$ It is deliberately not an integer, which is what rules out options A and C immediately.

3. 3

   Write the 501st term. $u_{501}=1+500d=1+500\cdot\frac{2017}{1000}=1+\frac{2017}{2}$

4. 4

   Combine into a single fraction. $u_{501}=\frac{2+2017}{2}=\frac{2019}{2}=1009.5$

5. 5

   Sanity-check with symmetry. Index $501$ is exactly halfway between $1$ and $1001$, so $u_{501}$ must be the average of the endpoints: $\frac{1+2018}{2}=\frac{2019}{2}$. Both routes agree, so the answer is B.