 # Find the sum of the arithmetic series 2 + 5 + 8 + ... + 56 excefebraxp 2022-09-14 Answered
Find the sum of the arithmetic series 2 + 5 + 8 + ... + 56
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The first term of the corresponding arithmetic sequence is a=2 and the common difference between successive terms is d=3.
Hence the general term may be given by the formula
$\left({x}_{n}\right)=a+\left(n-1\right)d$
$=2+\left(n-1\right)\left(3\right)$
$=3n-1$
So we now need to find which term has value 56 so that we know how many terms to sum up to.
$\therefore 56=3n-1⇒n=\frac{57}{3}=19$
So we thus need to find the sum of the series $\sum _{n=1}^{19}\left(3n-1\right)$
There are 2 formulae applicable :
1. ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$
$=\frac{19}{2}\left[2\left(2\right)+\left(19-1\right)\left(3\right)\right]$
$=551$
2. ${S}_{n}=\frac{n}{2}\left(a+l\right)$
$=\frac{19}{2}\left(2+56\right)$
$=551$

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