In an AP, if $S_{n} = n (4 n + 1)$ , fill the AP is 5, 13, __, ---

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Solution

$S_{n} = n (4 n + 1)$
$∴ S_{1} = a = 1 (4 + 1) = 5$
and $S_{2} = a_{1} + a_{2} = 2 (4 \times 2 + 1) = 18$
$\Rightarrow a + a + d = 18 \Rightarrow 2 a + d = 18$
$\Rightarrow d = 18 - 2 a = 18 - 10 = 8$
Therefore the AP is $a, a + d, a + 2 d, . . . .$
i.e. $5, 13, 21, . . .$

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