Question

Find the sum of the series of the arithmetic series:

7 + 13 + . . . + 601

a. 182,704
b. 60,800
c. 30,400
d. 15,200

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Answer 1

The answer would be d I think

Answer 2

`\bf 7~~,~~\stackrel{7+6}{13}......601\qquad \qquad \stackrel{\textit{common difference}}{d = 6} \\\\[-0.35em] ~\dotfill\\\\ 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}\\ \cline{1-1} a_1=7\\ d=6\\ a_n=601 \end{cases} \\\\\\ 601=7+(n-1)6\implies 601=7+6n-6\implies 601=1+6n \\\\\\ 600=6n\implies \cfrac{600}{6}=n\implies 100=n \\\\[-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}\\ \cline{1-1} a_1=7\\ a_n=601\\ n=100 \end{cases}\implies S_(100)=\cfrac{100(7+601)}{2} \\\\\\ S_(100)=\cfrac{60800}{2}\implies S_(100)=30400`