The sum of n,2n, 3n terms of an A.P are S1,S2, S3 . Prove that S3 = 3(

1.	The sum of n,2n, 3n terms of an A.P are S1,S2, S3 . Prove that S3 = 3(S2-S1)
Answer» S1= sum of n terms = n\|2{2a+(n-1)d}S2= sum of 2n terms = 2n\|2{2a+(2n-1)d}S3= sum of 3n terms= 3n\|2{2a+(3n-1)d}Lhs=3(S2-S3)=2n\|2 {2a+(2n-1)d} - n\|2 {2a+(n-1)d}=3n\|2{2a+(3n-1)d}Hence S3=3(S2-S1) \tLet the first term of the A.P. be\xa0a\xa0and the common difference be\xa0d.According the question,We have to prove that S3\xa0= 3 ( S2\xa0– S1).R.H.S = 3 (S2\xa0– S1)\xa0\xa0= S3\xa0= L.H.S⇒ S3\xa0= 3 ( S2\xa0– S1)\t\xa0

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