Show that the sequence `t_(n)` defined by `t_(n)=2*3^(n)+1` is not a GP.

Explanation

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