The sum of n, 2n , 3n terms of an AP are S1,S2 ,S3 respectively. Prove

1.	The sum of n, 2n , 3n terms of an AP are S1,S2 ,S3 respectively. Prove that S3 = 3(S2 -S1).
Answer» S1= n/2 [2a+(n-1) d]S2= 2n/2 [2a+(2n-1) d] S3= 3n/2 [2a+(3n-1) d] .......(i) S2-S1 = 2n/2 [2a +(2n-1) d] - n/2 [2a+(n-1) d ] = n/2[2a +(2n-1) 2d - 2a -(n-1) d ] = n/2 [2a + ( 4n-2-n+1) d ] = n/2 [ 2a + (3n-1) d ] (ie) 3 (S2-S1) = 3n/2 from (i) Hence proved . S1=a1S2 =a1+a2=n+2n=3nS3 =a1+a2+a3=n+2n+3n=6nLhs=S3=6nRhs=3(3n-n)Rhs=6nLhs=RhsHp It is not possible

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