In an AP, if sn =n (4n + 1),

We know that, the n th term of an AP is

$a_{n}=S_{n}-S_{n-1}$

$a_{n}=n(4 n+1)-(n-1)\{4(n-1)+1\} \quad\left[\because S_{n}=n(4 n+1)\right]$

$\Rightarrow \quad a_{n}=4 n^{2}+n-(n-1)(4 n-3)$

$=4 n^{2}+n-4 n^{2}+3 n+4 n-3=8 n-3$

Put $n=1, \quad a_{1}=8(1)-3=5$

Put $n=2, \quad a_{2}=8(2)-3=16-3=13$

Put $n=3, \quad a_{3}=8(3)-3=24-3=21$

Hence, the required AP is 5,13, 21,…