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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

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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)