The sum of first n, 2nn, 2n, and 3n3n terms of an AP is S1, S2S1, S2, and S3S3 respectively. Prove that S3=3(S2S1).S3=3(S2S1).


Answer:


Step by Step Explanation:
  1. We are told that

    S1S1 = Sum of first nn terms
    S2S2 = Sum of first 2n2n terms
    S3S3 = Sum of first 3n3n terms
  2. We know that the sum of first nn terms of an AP is given by Sn=n2(2a+(n1)d),Sn=n2(2a+(n1)d), where aa is the first term and nn is the number of terms in the AP.

    Therefore, we have S1=n2(2a+(n1)d)S2=2n2(2a+(2n1)d)S3=3n2(2a+(3n1)d)
  3. Now, 3(S2S1)=3(2n2(2a+(2n1)d)n2(2a+(n1)d))=3n2(2(2a+(2n1)d)(2a+(n1)d))=3n2(4a+4nd2d)(2a+ndd))=3n2(4a+4nd2d2and+d)=3n2(2a+3ndd)=3n2(2a+(3n1)d)=S3
  4. Thus, S3=3(S2S1).

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