Let the sum of n, 2n, …
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Sia ? 6 years, 1 month ago
Given: {tex}{S_1} = {n \over 2}\left[ {2a + (n - 1)d} \right]{/tex} …..(i)
{tex}{S_2} = {{2n} \over 2}\left[ {2a + (2n - 1)d} \right]{/tex}…..(ii)
And {tex}{S_3} = {{3n} \over 2}\left[ {2a + (3n - 1)d} \right]{/tex}
Now, {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} 3(S2 - S1) = S3
Hence proved.
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