-8 - 3 + 2 + 7 + ... + 67 hmm... so the first term's value is -8. If you notice, it went from -8, to -3 and then +2 and so on.... you can always get the "common difference" d by simply getting the difference of two terms, so d = +2 -(- 3) ----> d = +2+3 ----> d = 5.
now.. the last value is 67.. but what term is that anyway?
well, let's use the arithmetic sequence equation to see who that is.
[tex]\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
a_1=-8\\
d=5\\
a_n=67
\end{cases}
\\\\\\
a_n=-8+(n-1)5\implies 67=-8+(n-1)5
\\\\\\
67=-8+5n-5\implies 67=-13+5n\implies 80=5n
\\\\\\
\cfrac{80}{5}=n\implies \boxed{16=n}[/tex]
so, is the 16th term alrite.. .hmmm, so it begins with -8, then keeps on adding 5 16 times
[tex]\bf \sum\limits_{n=1}^{16}~-8+(n-1)5\implies \sum\limits_{n=1}^{16}~5n-13[/tex]
both forms are the same, so.. hmm it depends on how simplified you need it.
