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

1 Answer
Mar 28, 2016

Answer:

#sum_(n=1)^19(3n-1)=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

#(x_n)=a+(n-1)d#

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

#=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=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 :

  1. #S_n=n/2[2a+(n-1)d]#

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

#=551#

  1. #S_n=n/2(a+l)#

#=19/2(2+56)#

#=551#