To find the first term, we note that the $\var{n}$^th term is $\simplify{{t1}+({n}-1)*{d}}$ and the common difference is $\var{d}$ hence we can calculate the first term as $\simplify{{t1}+({n}-1)*{d}}-(\var{n}-1)\times\var{d}$

To find the sum of the first $\var{tsum}$ terms we used the known formula $S_n = \frac{n(a_1 + a_n)}{2}$ where $a_i$ is the $i$^th term of the arithmetic series. Since we know how to calculate $a_n$ from our above considerations, we know that $S_n = \frac{n(a_1+a_1+(n-1)\times d)}{2}$  which simplifies to $S_n = \frac{n(2\times\var{t1}+(n-1)\var{d})}{2}$