In an AP, If `S_(n)=n(4n+1),` then find the AP.

Explanation

We know that, the `n`th term of an AP is
`a_(n)=S_(n)-S_(n-1)`
`a_(n)=n(4n+1)-(n-1){4(n-1)+1} " " [:' S_(n)=n(4n+1)]`
`a_(n)=4n^(2)+n-(n-1)(4n-3)`
` " "=4n^(2)+n-4n^(2)+3n+4n-3=8n-3`
Put `n=1, " " a_(1)=8(1)-3=5`
Put `n=2, " " a_(2)=8(2)-3=16-3=13`
Put `n=3, " " a_(3)=8(3)-3=24-3=21`
Hence, the required AP is 5, 13, 21, ...