If $S_{n}$ denotes the sum of $n$ terms of an $A P$ whose common difference is $d$ and first term is $a$ .
Find $S_{n} - 2 S_{n} + S_{n + 2}$ .

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Solution

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$S_{n} + 2 = \frac{(n + 2)}{2} (2 a + (n + 2 - 1) d]$
$= \frac{n + 2}{2} (2 a + (n + 1) d)$
To find
$S_{n} - 2 S_{n} + S_{n + 2} = - S_{n} + S_{n + 2}$
$= (- \frac{n}{2}) [2 a + (n - 1) d] + (\frac{n + 2}{2}) [2 a + (n + 1) d]$
$= - \frac{2 a n}{2} - \frac{n^{2} d}{2} + \frac{n d}{2} + \frac{2 a n}{2} + \frac{n^{2} d}{2} + \frac{n d}{2} + 2 a + n d + d$
$= \frac{2 n d}{2} + 2 a + n d + d$
$= 2 n d + 2 a + d$ .

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If Sn denotes the sum of n terms of an AP whose common difference is d and first term is a.Find Sn−2Sn+Sn+2.

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If $S_{n}$ denotes the sum of $n$ terms of an $A P$ whose common difference is $d$ and first term is $a$ .
Find $S_{n} - 2 S_{n} + S_{n + 2}$ .