Part a)

To find the first term $a_1$, we need to first start with the formula for the $n^{th}$ term:

$a_n=a_1+d(n-1)$ $_{..(i)}$    which is rearranged in terms of $a_1$:    $a_1=a_n-d(n-1)$ $_{..(ii)}$

We then identify the variables which we know, namely: $n=\var{n}$,    $a_n=\simplify{{t1}+({n}-1)*{d}}$,    $d=\var{d}$

These values can then be subsituted into $_{(ii)}$ and solved for $a_1$:

$a_1=\simplify{{t1}+({n}-1)*{d}}-\var{d}(\var{n}-1)$

Part b)

To find the sum of the first $\var{tsum}$ terms we take the formula for the sum of an arithmetic series:

$S_n=\frac{n}{2}(a_1+a_n)$ $_{..(iii)}$

While we don't yet know the value of $a_n$, it is obtainable with formula $_{..(i)}$. It is common to substitute this formula into the formula for the sum as follows:

$S_n=\frac{n}{2}(a_1+a_1+d(n-1))$

$\therefore\;\;\;S_n=\frac{n}{2}(2a_1+d(n-1))$    $_{..(iv)}$

We identify the variables which we know, namely: $n=\var{tsum}$,    $a_1=\var{t1}$,    $d=\var{d}$

They are substituted into $_{..(iv)}$ and solved for $S_n$:

$S_\var{tsum}=\frac{\var{tsum}}{2}(2\times\var{t1}+\var{d}(\var{tsum}-1))$

Part c)

To determine the least value of $n$ for which the sum of the first $n$ terms of the series exceeds $1000$, the sum formula $_{..(iv)}$ is set up with the lower limit value of $1000$ substituted in:

$1000=\frac{n}{2}(2\times\var{t1}+\var{d}(n-1))$

It is then expanded algebraically to give the following quadratic:

$\simplify{{d}/2}n^2+\simplify{{2*{t1}-{d}}/2}n-1000=0$

$n$ can then be found using the quadratic formula:

$n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$    where $a=\simplify{{d}/2}$,    $b=\simplify{{2*{t1}-{d}}/2}$   and $c=-1000$

Thus,

$n=\frac{-\var{b}\pm\sqrt{(\var{b})^2-\simplify{{-1000*4*{a}}}}}{\simplify{2*{a}}}$

$\therefore\;\;\;n=-\left(\simplify{{2*{t1}-{d}}/{2*2*{a}}}\right)\pm\left(\frac{\sqrt{\simplify{{bsq}-{-1000*4*{a}}}}}{\simplify{2*{a}}}\right)$

In this context, we are talking of positive $n^{th}$ terms, so we disregard the negative square root.

The result is then computed to be: $\var{nc}$

If this is the value of $n$ which produces $1000$, it follows that the next whole number will be the least value of $n$ for which the sum of the first $n$ terms of the series exceeds $1000$.