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

Question

4164 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-28T10:35:55

$19 \sum n = 1 (3 n - 1) = 551$ $sum_(n=1)^19(3n-1)=551$

The first term of the corresponding arithmetic sequence is $a = 2$ $a=2$ and the common difference between successive terms is $d = 3$ $d=3$ .

Hence the general term may be given by the formula

$(x_{n}) = a + (n - 1) d$ $(x_n)=a+(n-1)d$

$= 2 + (n - 1) (3)$ $=2+(n-1)(3)$

$= 3 n - 1$ $=3n-1$

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

$therefore 56=3n-1 => n=57/3=19$

So we thus need to find the sum of the series $sum_(n=1)^19(3n-1)$ .

There are 2 formulae applicable :

$=19/2[2(2)+(19-1)(3)]$

$=551$

$=19/2(2+56)$

$=551$