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Solution :

We have, `t_(n)=5n+4` <br> On replacing n by `(n-1)`, we get <br> `t_(n-1)=5(n-1)+4` <br> `implies t_(n-1)=5n-1` <br> `therefore t_n-t_(n-1)=(5n+4)-(5n-1)=5` <br> Clearly, `t_n-t_(n-1)` is independent of n and is equal to 5. So, the given sequence is an AP with common difference 5.