Part a)
To find the first term a1, we need to first start with the formula for the nth term:
an=a1+d(n−1) ..(i) which is rearranged in terms of a1: a1=an−d(n−1) ..(ii)
We then identify the variables which we know, namely: n=14, an=32, d=2
These values can then be subsituted into (ii) and solved for a1:
a1=32−2(14−1)
Part b)
To find the sum of the first 13 terms we take the formula for the sum of an arithmetic series:
Sn=n2(a1+an) ..(iii)
While we don't yet know the value of an, it is obtainable with formula ..(i). It is common to substitute this formula into the formula for the sum as follows:
Sn=n2(a1+a1+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.