Question

if the sum of first n,2n,3n terms of an A.P. be s1,s2,s3 then prove that s3=3(s2-s1)

Answer 1

lets we start the question from given
n= s1
2n= s2
3n= s3
to prove= s3=3(s2-s1)
l.h.s.===s3=3n
r.h.s===3(2n-n)
=3n
so it prove that l.h.s = r.h..s

Answer 2

Sn of A.P.is given by Sn = n/2[2*a + (n-1)*d] where n= no.of terms , a= first term & d=common difference. 
So S1 = Sn = n/2*[2*a + (n-1)*d] . for S2 replace '2n' for 'n' 
Now S2 = S2n = n*[2*a + (2n-1)*d] for S3 replace '3n' for 'n' 
Now S3 = S3n = 3n/2*[2*a + (3n-1)*d] ---------------- [1] 
Hence S2 - S1 = n*[2*a + (2n-1)*d] - n/2*[2*a + (n-1)*d] simplifies to n/2*(2*a+3*n*d-d) 
= n/2*[2*a +(3n-1)*d] ----- [2] 
Now [1]& [2] clearly show that S3=3*(S2-S1)