Let the sum of n, 2n, 3n, terms of an A.P be s1,s2,s3,

1.	Let the sum of n, 2n, 3n, terms of an A.P be s1,s2,s3,
Answer» Given: {tex}{S_1} = {n \\over 2}\\left[ {2a + (n - 1)d} \\right]{/tex}\xa0…..(i){tex}{S_2} = {{2n} \\over 2}\\left[ {2a + (2n - 1)d} \\right]{/tex}…..(ii)And\xa0{tex}{S_3} = {{3n} \\over 2}\\left[ {2a + (3n - 1)d} \\right]{/tex}Now,\xa0{tex}{S_2} - {S_1} = {{2n} \\over 2}\\left[ {2a + (2n - 1)d} \\right] - {n \\over 2}\\left[ {2a + (n - 1)d} \\right]{/tex}{tex}\\Rightarrow {S_2} - {S_1} = (n - {n \\over 2})2a + \\left[ {n(2n - 1) - {n \\over 2}(n - 1)} \\right]d{/tex}{tex}= na + {1 \\over 2}\\left[ {4{n^2} - 2n - {n^2} + n} \\right]d{/tex}{tex}= {n \\over 2}\\left[ {2a + (3n - 1)d} \\right] = {1 \\over 3}\\left\\{ {{{3n} \\over 2}\\left[ {2a + (3n - 1)d} \\right]} \\right\\} = {1 \\over 3}{S_3}{/tex}{tex}\\Rightarrow {/tex}\xa03(S2\xa0- S1) = S3Hence proved.

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