s[2m]=2m[a 1]+2m(2m- 1)d/2=2m[a 1]+(4m^2-2m)d/2

s[3m]=3m[a 1]+3m(3m- 1)d/2=3m[a 1]+(9m^2-3m)d/2

s[2m]-s[m]=m[a 1]+(3m^2-m)d/2=m[a 1]+(m^2-m)d/2+m^2*d=s[m]+m^2*d

s[3m]-s[2m]=m[a 1]+(5m^2-m)d/2=m[a 1]+(m^2-m)d/2+2m^2*d=s[m]+2m^2*d

So S[m], S[2m]-S[m], S[3m]-S[2m] become arithmetic progression.

2(S[2m]-S[m])=S[m]+S[3m]-S[2m]

S[3m]= 3(S[2m]-S[m])= 3( 100-30)= 2 10

The solution of the high-powder answer zzz680 13 1 is correct, but S[m], S[2m]-S[m] are omitted as proof of arithmetic progression, and the subject may be confused.