1.

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

Answer»

Given:

an = 2/3n

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

So,

a1 = 2/3

a2 = 2/32 = 2/9

a3 = 2/33 = 2/27

a4 = 2/34 = 2/81

In GP,

a3/a2 = (2/27) / (2/9)

= 2/27 × 9/2

= 1/3

a2/a1 = (2/9) / (2/3)

= 2/9 × 3/2

= 1/3

∴ Common ratio of consecutive term is 1/3. Hence n ∈ N is a G.P.



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