Find an A.P. in which the sum of any number of terms is always three times the squared number of these terms.

Given:

$S_{n}=3 n^{2}$

For $n=1, S_{1}=3 \times 1^{2}=3$

For $n=2, S_{2}=3 \times 2^{2}=12$

For $n=3, S_{3}=3 \times 3^{2}=27$

and so on

$\therefore S_{1}=a_{1}=3$

$a_{2}=S_{2}-S_{1}=12-3=9$

$a_{3}=S_{3}-S_{2}=27-12=15$

and so on

Thus, the A.P. is $3,9,15 \ldots$