Please tell

sum of n terms of an A.P. with first term a, common difference d, is

S = n/2 ( 2a+(n-1) d)

so

S1 = n/2 (2+(n-1))

S2 = n/2 (2+(n-1)2) = n²

S3 = n/2 (2+(n-1)3)

so S1+S3 = n/2 ( 4+4(n-1) ) = 2n² = 2S2

hence proved