Show that the sequence defined by `a_n=2/(3^n),\ n in N\ ` is a G.P.

Explanation

To show that the sequence defined by \( a_n = \frac{2}{3^n} \) for \( n \in \mathbb{N} \) is a geometric progression (G.P.), we need to demonstrate that the ratio of consecutive terms is constant.

### Step-by-Step Solution:

1. **Define the terms of the sequence**:
The \( n \)-th term of the sequence is given by:
\[
a_n = \frac{2}{3^n}
...