Answer:
Sum of Arithmetic Sequence
[tex]S_{n} = (a-l)[/tex] or [tex]S_{n} = \frac{1}{2} n(2a + (n-1)d)[/tex]
Where:
a = -3
l is unknown
d = 1 - -3 = 4
n = 40
Use the second equation because l in unknown.
[tex]S_{40} = \frac{1}{2}40(2(-3) + (40-1)4)[/tex]
[tex]S_{40} = \frac{1}{2}40(-6 + (39)4)\\S_{40} = \frac{1}{2}40(-6 + 156)\\S_{40} = \frac{1}{2}40(150)\\S_{40} = \frac{1}{2}6000\\S_{40} = 3000[/tex]
[tex]\bf -3~~,~~\stackrel{-3+4}{1}~~,~~\stackrel{1+4}{5}~~,~~...\qquad \qquad \stackrel{\textit{common difference}}{d = 4} \\\\\\ n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ d=\textit{common difference}\\[-0.5em] \hrulefill\\ a_1=-3\\ d= 4\\ n= 40 \end{cases} \\\\\\ a_{40}=-3+(40-1)4\implies a_{40}=-3+(39)4\implies a_{40}=153 \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf \textit{sum of a finite arithmetic sequence} \\\\ S_n=\cfrac{n(a_1+a_n)}{2}\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\[-0.5em] \hrulefill\\ n = 40\\ a_{40}=153\\ a_1=-3 \end{cases}\implies S_{40}=\cfrac{40(-3+153)}{2} \\\\\\ S_{40}=20(150)\implies S_{40}=3000[/tex]
