Write the sum $LaTeX: \displaystyle 45+51+57 \ldots +207+213$ in sigma notation and then find the sum.

The common difference is given by $LaTeX: \displaystyle a_2-a_1=51-(45)=6$ . Using the first term gives the sequene $LaTeX: \displaystyle a_n= 45+(n-1)(6)$ . Setting the general term equal to the last term and solving for $LaTeX: \displaystyle n$ gives $LaTeX: \displaystyle 45+(n-1)(6)=213 \implies n = 29$ . Writing in sigma notation gives $LaTeX: \displaystyle \displaystyle \sum_{n=1}^{29} \left(6 n + 39\right)$ . Using the formula for a finite arithmetic sum gives $LaTeX: \displaystyle \frac{ 29(45+213) }{2}=3741$ .