If the sum of n, 2n and 3n terms of an AP be S1, S2 and S3 respectively then prove that S3=3(S2-S1).

If the sum of n, 2n and 3n terms of an AP be S1, S2 and S3 respectivel

1.	If the sum of n, 2n and 3n terms of an AP be S1, S2 and S3 respectively then prove that S3=3(S2-S1).
Answer» Let a be the first term and d be the common difference of the given AP. Then,S1 = sum of first n terms of the given AP,S2 = sum of first 2n terms of the given AP,S3 = sum of first 3n terms of the given AP.S1 ={tex}\\frac{n}{2}{/tex}{tex}\\cdot{/tex}{2a+(n-1)d}, S2={tex}\\frac{{2n}}{2}{/tex}{tex}\\cdot{/tex}{2a+(2n-1)d}, and S3= {tex}\\frac{{3n}}{2}{/tex}{tex}\\cdot{/tex}{2a+(3n-1)d}{tex}\\Rightarrow{/tex}3(S2-S1) = 3{tex}\\cdot{/tex}[{2na+n(2n - 1)d}\xa0- {na+{tex}\\frac{1}{2}{/tex}n(n-1)d}]= 3{tex}\\cdot{/tex}[na + {tex}\\frac{3}{2}{/tex}n2d-{tex}\\frac{1}{2}{/tex}nd] = {tex}\\frac{{3n}}{2}{/tex}{tex}\\cdot{/tex}[2a+3nd - d]= {tex}\\frac{{3n}}{2}{/tex}{tex}\\cdot{/tex}[2a+(3n-1)d}=S3.Hence, S3=3(S2-S1).