Write the sum $LaTeX: \displaystyle 48+56+64 \ldots +216+224$ in sigma notation and then find the sum.

The common difference is given by $LaTeX: \displaystyle a_2-a_1=56-(48)=8$ . Using the first term gives the sequene $LaTeX: \displaystyle a_n= 48+(n-1)(8)$ . Setting the general term equal to the last term and solving for $LaTeX: \displaystyle n$ gives $LaTeX: \displaystyle 48+(n-1)(8)=224 \implies n = 23$ . Writing in sigma notation gives $LaTeX: \displaystyle \displaystyle \sum_{n=1}^{23} \left(8 n + 40\right)$ . Using the formula for a finite arithmetic sum gives $LaTeX: \displaystyle \frac{ 23(48+224) }{2}=3128$ .