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Solution :

We have, `t_n=3n^(2)+2` <br> On replacing n by `(n-1),` we get <br> `t_(n-1)=3(n-1)^(2)+2` <br> `implies t_(n-1)=3n^(2)-6n+5` <br> `therefore t_n-t_(n-1)=(3n^(2)+2)-(3n^(2)-6n+5)` <br> `" "=6n-3` <br> Clearly, `t_n-t_(n-1)` is not independent of n and therefore it is not constant. So, the given sequence is not an AP.