Question

Q.18

Answer 1

`S_n=an^2+bn`

Replaing n by (n-1), we get 

`S_(n-1)=a(n-1)^2+b(n-1)`

         `=a(n^2-2n+1)+(bn-b)`

        ` = an^2-2an+a+bn-b`

`∴ t_n=S_n-S_(n-1)`

    ` =(an^2+bn)-(an^2-2an+a+bn-b)`

    `=an^2+bn-an^2+2an-a-bn+b`

     `=2an-a+b`

Replacing n by (n-1), we get 

`t_(n-1)=2a(n-1)-a+b`

          `=2an-2a-a+b`

Now, 

`t_n-t_(n-1)=(2an-a+b)-(2an-2a-a+b)`

                  `=2an-a+b-2an+2a+a-b`

                  =2a, Which is constant, independet of n 

Thus, the sequence is an A.P.