If the sum of first n,2n and 3n terms of Ap be s1,s2&s3 respectively than prove that s3=3 (s2-s1)

If the sum of first n,2n and 3n terms of Ap be s1,s2&s3 respectively t

1.	If the sum of first n,2n and 3n terms of Ap be s1,s2&s3 respectively than 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).