Show that the sequence defined by $$a_n = \dfrac{2}{3}^n$$, n ∈ N is a G.P.

Asked by Sakshi | 1 year ago |  110

##### Solution :-

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.

Answered by Sakshi | 1 year ago

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