If $S_{1}, S_{2} a n d S_{3}$ denote the sum of first $n_{1}, n_{2} a n d n_{3}$ terms respectively of an AP, then $\frac{S_{1}}{n_{1}} (n_{2} - n_{3}) + \frac{S_{2}}{n_{2}} (n_{3} - n_{1}) + \frac{S_{3}}{n_{3}} (n_{1} - n_{2})$ is equal to

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$S_{1} S_{2} S_{3}$

$n_{1} n_{2} n_{3}$

answer is A.

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Detailed Solution

We have,

$S_{1} = \frac{n_{1}}{2} [2 a + (n_{1} - 1) d] \Rightarrow \frac{2 S_{1}}{n_{1}} = 2 a + (n_{1} - 1) d$

$S_{2} = \frac{n_{2}}{2} [2 a + (n_{2} - 1) d] \Rightarrow \frac{2 S_{2}}{n_{2}} = 2 a + (n_{2} - 1) d$

$S_{3} = \frac{n_{3}}{2} [2 a + (n_{3} - 1) d] \Rightarrow \frac{2 S_{3}}{n_{3}} = 2 a + (n_{3} - 1) d$

$n o w \frac{2 S_{1}}{n_{1}} (n_{2} - n_{3}) + \frac{2 S_{2}}{n_{2}} (n_{3} - n_{1}) + \frac{2 S_{3}}{n_{3}} (n_{1} - n_{2})$

$= [2 a + (n_{1} - 1) d] (n_{2} - n_{3}) + [2 a + (n_{2} - 1) d] (n_{3} - n_{1})$ $+ [2 a + (n_{3} - 1) d] (n_{1} - n_{2})$

$= (2 a - d) [(n_{2} - n_{3}) + (n_{3} - n_{1}) + (n_{1} - n_{2})] + d (n_{1} [n_{2} - n_{3}] + n_{2} [n_{3} - n_{1}] + n_{3} [n_{1} - n_{2}])$

$= 0$

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