If S_1, S_2 and S_3 are the sums of the first | vedclass

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(A) Let the first term be $a$ and the common difference be $d$.The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Thus,$\frac

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(A) Let the first term be $a$ and the common difference be $d$.The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Thus,$\frac{S_n}{n} = a + \frac{(n-1)d}{2}$.Substituting this into the expression:$\frac{S_1}{n_1}(n_2 - n_3) + \frac{S_2}{n_2}(n_3 - n_1) + \frac{S_3}{n_3}(n_1 - n_2)$$= [a + \frac{(n_1-1)d}{2}](n_2 - n_3) + [a + \frac{(n_2-1)d}{2}](n_3 - n_1) + [a + \frac{(n_3-1)d}{2}](n_1 - n_2)$$= a(n_2 - n_3 + n_3 - n_1 + n_1 - n_2) + \frac{d}{2}[(n_1-1)(n_2-n_3) + (n_2-1)(n_3-n_1) + (n_3-1)(n_1-n_2)]$$= a(0) + \frac{d}{2}[n_1n_2 - n_1n_3 - n_2 + n_3 + n_2n_3 - n_2n_1 - n_3 + n_1 + n_3n_1 - n_3n_2 - n_1 + n_2]$$= 0 + \frac{d}{2}(0) = 0$.

If $S_1, S_2$ and $S_3$ are the sums of the first $n_1, n_2$ and $n_3$ terms of an arithmetic progression respectively,then $\frac{S_1}{n_1}(n_2 - n_3) + \frac{S_2}{n_2}(n_3 - n_1) + \frac{S_3}{n_3}(n_1 - n_2) = ....$

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