Write the sum $LaTeX: \displaystyle 40+46+52 \ldots +328+334$ in sigma notation and then find the sum.

The common difference is given by $LaTeX: \displaystyle a_2-a_1=46-(40)=6$ . Using the first term gives the sequene $LaTeX: \displaystyle a_n= 40+(n-1)(6)$ . Setting the general term equal to the last term and solving for $LaTeX: \displaystyle n$ gives $LaTeX: \displaystyle 40+(n-1)(6)=334 \implies n = 50$ . Writing in sigma notation gives $LaTeX: \displaystyle \displaystyle \sum_{n=1}^{50} \left(6 n + 34\right)$ . Using the formula for a finite arithmetic sum gives $LaTeX: \displaystyle \frac{ 50(40+334) }{2}=9350$ .