Question 26498
The 1st term is a(1)=1
d is the difference between any two consecutive terms: e.g. d=3-1=2
The nth term is a(n)=a(1)+(n-1)d
The sum of n terms is S(n)=(1/2)n[a(1)+a(n)]
101st term: a(101)=a(1)+(100)d
                  =1+100*2=201
Sum of 1st 20 terms:
a(20)=1+(19)2=1+38=39
S(20)=(1/2)(20)(1+39)
     =10(40)=400
Sum of 1st 30 terms:
a(30)=a(1)+29(2)
     =1+58=59
S(30)=(1/2)(30)(1+59)
     =15(60)
     =900
Cheers,
stan H.