To find the common difference d we note that the $\var{n1}$^th term is $\var{t1}$ and the $\var{n2}$^th term is $\var{t2}$ so we can find the common difference d by noting $\var{t2}-\var{t1}=(\var{n2}-\var{n1})\times d$, hence common difference $= \var{d}$

To find the least positive value for n for which the sum of the first n terms of the series exceeds 1000, we recall that the sum of an arithmetic series can be found with the formula $S_n = \frac{n(2a_1 + (n-1)\times d)}{2}$ where $a_1$ denotes the first term and d denotes the common difference. Now we want to find the smallest positive $n$ such that $S_n > 1000$. This is now a matter of solving a quadratic, which is left as an exercise.