Question

Find the sum of the first 40 terms of the arithmetic sequence-3,1,5,...

Answer 1

`\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`

`\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`

Answer 2

Sum of Arithmetic Sequence

`S_(n) = (a-l)` or
`S_(n) = (1)/(2) n(2a + (n-1)d)`

Where:

• a = first term
• l = last term
• d = common difference
• n = number of terms

a = -3

l is unknown

d = 1 - -3 = 4

n = 40

Use the second equation because l in unknown.

`S_(40) = (1)/(2)40(2(-3) + (40-1)4)`

`S_(40) = (1)/(2)40(-6 + (39)4)\\S_(40) = (1)/(2)40(-6 + 156)\\S_(40) = (1)/(2)40(150)\\S_(40) = (1)/(2)6000\\S_(40) = 3000`