A2. What is $1+3.5+6+8.5+\ldots+2501$, if the difference between consecutive terms is constant?
(A) 1249750
(B) 1251000
(C) 1251750.5
(D) 1252251
(E) 1253502.5
A2. We find that we need to sum $n$ terms of an arithmetic sequence $\mathrm{z}$ with a difference $d=2.5$. Considering $a_{n}=a_{1}+(n-1) d$ or $2501=1+(n-1) \cdot 2.5$. Solving the equation, we find that we need to sum 1001 terms. Considering $S_{n}=\frac{n}{2}\left(a_{1}+a_{n}\right)$ or $S_{n}=\frac{n}{2}\left(2 a_{1}+(n-1) d\right)$, we get $S_{1001}=1252251$.