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Solution :

We have, `t_(n)=2*3^(n)+1` <br> On replacing n by `(n-1)` in `t_(n)`, we get <br> `t_(n-1)=2*3^(n-1)+1` <br> `implies t_(n-1)=(2*3^(n)+3)/(3)` <br> `therefore (t_(n))/(t_(n-1))=((2*3^(n)+1))/(((2*3^(n)+3))/(3))=(3(2*3^(n)+1))/((2*3^(n)+3))` <br> Clearly, `(t_(n))/(t_(n-1))` is not independent of n and is therefore not constant. So, the given sequence is not a GP.