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Let the sum of n ,2n,3n …

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Let the sum of n ,2n,3n terms of an AP be S1,S2 and S3 show that S3 =3(S2-S1)?
  • 1 answers

Abhishek Saini 7 years, 2 months ago

Let a be the first term and d be the comman ratio. S1 = n/2[ 2a+(n-1)d ] S2 = 2n/2[ 2a+(3n-1)d ] S3 = 3n/2[ 2a+(3n-1)d ] Now, Right Hand Side (RHS) = 3( S2- S1) = 3 { 2n/2[2a+(2n-1)d] - n/2[2a+(n-1)d]} On solving this , we get = 3{ an + 3n²d/2 -nd/2 } = 3 { n [ a + ( 3n-1)d/2 ] } = 3n/2 [ 2a + ( 3n-1) ] = S3 .
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