You could first note that the last (20th) term added will be #22-5*19=-73# (think about why this makes sense) and write your summation as #S=22+17+12+7+2+\cdots+(-63)+(-68)+(-73)#. Now write #S=-73+(-68)+(-63)+(-58)+(-53)\cdots+12+17+22# directly beneath the first summation and add these equations to get #2S=20*(-51)=-1020# (there are 20 terms that each add to #-51#). Now divide by 2 to get #S=-510#.
Alternatively, you could recognize the summation as an arithmetic series with #n=20# terms, first term #a_{1}=22#, and "common difference" #d=17-22=-5#. Then use the formula #S_{n}=(n/2)*(2a_{1}+(n-1)d)# to get
#S_{20}=(20/2)*(2*22+19*(-5))=10*(44-95)#
#=10*(-51)=-510#.
