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

Answer:

Sum of Arithmetic Sequence

$S_{n} = (a-l)$ or $S_{n} = \frac{1}{2} n(2a + (n-1)d)$

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.

$S_{40} = \frac{1}{2}40(2(-3) + (40-1)4)$

$S_{40} = \frac{1}{2}40(-6 + (39)4)\\S_{40} = \frac{1}{2}40(-6 + 156)\\S_{40} = \frac{1}{2}40(150)\\S_{40} = \frac{1}{2}6000\\S_{40} = 3000$