The sum of the first n natural numbers is given by
S(n)=n(n+1)2.
∴S(15)−S(11)=15(15+1)2−11(11+1)2
=120−66
=54
Similarly, S(22)−S(19)=22(22+1)2−19(19+1)2
=253−190
=63
S(17)−S(13)=17(17+1)2−13(13+1)2
=153−91
=62
S(11)−S(3)=11(11+1)2−3(3+1)2
=66−6
=60
Hence, S(22)−S(19)>S(17)−S(13)>S(11)−S(3)>S(15)−S(11).