The sum of 'n' terms of a progression is `(n^(2)+5n).` Prove that it is arithmetic progression. Also find its common difference.

Answer

N/a

Explanation

`"Here "S_(n)=n^(2)+5n`
`rArr " "S_(n-1)=(n-1)^(2)+5(n-1)`
`=n^(2)-2n+1+5n-5`
`=n^(2)+3n-4`
`therefore" "T_(n)=S_(n)-S_(n-1)`
`=(n^(2)+5n)-(n^(2)+3n-4)`
`=2n+4`
`rArr " "T_(n-1)=2(n-1)+4`
`=2n-2+4`
`=2n+2`
`"Now, "T_(n)-T_(n-1)=(2n+4)-(2n+2)=2`
which does not depend on 'n'.
So, the given progression is A.P. and its common difference is 2.