Question

Given the Arithmetic series A1+A2+A3+A4 11+15+19+23+...+71 what is the value of sum?

Answer 1

first off, let's notice the arithmetic sequence is moving as

$\bf 11~~,~~\stackrel{11+4}{15}~~,~~\stackrel{15+4}{19}~~,~~\stackrel{19+4}{23}...71$

so is adding "4" to the current term in order to get the next term, meaning the "common difference" of d = 4, and the first term is of course 11.

so, hmmm what ordinal value is the last number of 71 anyway?

$\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ d=4\\ a_1=11 \end{cases} \\\\\\ 71=11+(n-1)4\implies 71=11+4n-4\implies 71=7+4n \\\\\\ 64=4n\implies \cfrac{64}{4}=n\implies 16=n$

low and behold, so 71 is the 16th term, well, let's get their sum then.

$\bf \textit{sum of a finite arithmetic sequence} \\\\ S_n=\cfrac{n(a_1+a_n)}{2}\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ ----------\\ n=16\\ a_1=11\\ a_n=71 \end{cases} \\\\\\ S_{16}=\cfrac{16(11+71)}{2}\implies S_{16}=8(82)\implies S_{16}=656$

Answer 2

656 is the final answer :)