Show that the sequence defined by an = 2/3n, n ∈ N is a G.P.

Let us consider n = 1, 2, 3, 4, … since n is a natural number.

So,

a₁ = \( \dfrac{2}{3}\)

a₂ = \( \dfrac{2}{9}\)

a₃ = \( \dfrac{2}{27}\)

a₄ = \( \dfrac{2}{81}\)

In GP,

\(\dfrac{a_3}{a_2} = \dfrac{(\dfrac{2}{27}) }{ (\dfrac{2}{9})}\)

= \( \dfrac{1}{3}\)

\( \dfrac{a_2}{a_1} = \dfrac{(\dfrac{2}{9}) }{ (\dfrac{2}{3})}\)

 = \( \dfrac{1}{3}\)

Common ratio of consecutive term is \( \dfrac{1}{3}\). Hence n ∈ N is a G.P.