If the sum of first $2 n, 3 n$ and $5 n$ terms of an $A P$ be $S_{2}, S_{3}$ and $S_{5}$ respectively, the find the value of $S_{3} - S_{2}$ in terms off $S_{5}$ .

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Solution

$S_{2} = n (2 a + (2 n - 1) d)$
$S_{3} = \frac{3 n}{2} (2 a + (3 n - 1) d)$
$S_{5} = \frac{5 n}{2} (2 a + (5 n - 1) d)$
$S_{3} - S_{5} = n [\frac{3}{2} (2 a + 3 n d - d) - (2 a + 2 n d - d)]$
$= n [3 a + \frac{9}{2} n d - \frac{3}{2} d - 2 a - 2 n d + d]$
$= n [a + \frac{5 n d}{2} - \frac{d}{2}]$
$= \frac{n}{2} [2 a + (5 n - 1) d]$
$= \frac{1}{5} S_{5}$
$∴$ $S_{3} - S_{2} = \frac{S_{5}}{5}$

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