Show that the sequence `t_(n)` defined by `t_(n)=(2^(2n-1))/(3)` for all values of `n in N` is a GP. Also, find its common ratio.

Explanation

We have, `t_(n)=(2^(2n-1))/(3)`
On replacing n by `n-1`, we get
`t_(n-1)=(2^(2n-3))/(3) implies (t_(n))/(t_(n-1))=((2^(2n-1))/(3))/((2^(2n-3))/(3))=2^(2)=4 `
Clearly, `(t_(n))/(t_(n-1))` is independent of n and is equal to 4. So, the given sequence is a GP with common ratio 4.