# What is the sum of the arithmetic series 2 + 5 + 8 + ... + 56?

Mar 28, 2016

${\sum}_{n = 1}^{19} \left(3 n - 1\right) = 551$

#### Explanation:

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

$\left({x}_{n}\right) = a + \left(n - 1\right) d$

$= 2 + \left(n - 1\right) \left(3\right)$

$= 3 n - 1$

So we now need to find which term has value $56$ so that we know how many terms to sum up to.

$\therefore 56 = 3 n - 1 \implies n = \frac{57}{3} = 19$

So we thus need to find the sum of the series ${\sum}_{n = 1}^{19} \left(3 n - 1\right)$.

There are 2 formulae applicable :

1. ${S}_{n} = \frac{n}{2} \left[2 a + \left(n - 1\right) d\right]$

$= \frac{19}{2} \left[2 \left(2\right) + \left(19 - 1\right) \left(3\right)\right]$

$= 551$

1. ${S}_{n} = \frac{n}{2} \left(a + l\right)$

$= \frac{19}{2} \left(2 + 56\right)$

$= 551$