The arithmetic sequence is 13, 21, 29, 37... Find the least number of terms required for the sum of the sequence terms to exceed 1000. All I know is that a=13 and common difference=8. I am unaware how to continue from here.

Just to do it slightly differently.
If I have $a_0 = 13$ and $a_k = 13 + 8*k$ and I have a list of $n$ of these terms $a_0=13, a_1=13+8, a_2=13+2*8,......, a_n=13 + n*8$, what is the average value of all of them.
Well as they are all exactly $8$ apart from each other, the average value will be directly in the middle  or $13 + \frac n2*8$.
So what do you get when you add up all the values?  Well you nave $(n+1)$ terms.  And the average term is $13+\frac n2*8$.  So when you add them all up you should get $(n+1)(13 + \frac n2*8)$.
So $(n+1)(13+\frac n2*8) > 1000$
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But one thing you should teach yourself.  $1 + 2 + 3 + 4 + 5 + ...... +n=???$.
Well the average is $\frac {n+1}2$.  ANd there are $n$ terms so $???? = \frac {n(n+1)}2$.  You will see that formula a lot!  
Alternative proof:
$1 + 2 + 3 + 4 + .......... + n  = S$
$n + (n-1) + (n-2) + (n-3) + ... + 1 = S$
$(n+1) + (n-1+2) + (n-2+3) + (n-3 + 4) + ..........+(1+n) = 2S$
$(n+1) + (n+1)+ (n+1) + (n+1) + ...... + (1+n) = 2S$
$n(n+1) = 2S$
$\frac {n(n+1)}2 = S$.
.....
So we have $13 + (13 + 8) + (13+2*8) + (13+2*8) + ......  + (13+n*8) = $
$[13 + 13 + 13 + ....... + ] + [8 + 2*8 + 3*8 + ........ + n*8]=$
$(n+1)13 + 8[1 + 2 + 3 + ........ + n] =$
$(n+1)13 + 8*\frac {n(n+1)}2 = $
$(n+1)(13 + 4n) > 1000$.
