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

So,

a_{1} = \( \dfrac{2}{3}\)

a_{2} = \( \dfrac{2}{9}\)

a_{3} = \( \dfrac{2}{27}\)

a_{4} = \( \dfrac{2}{81}\)

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 agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Find the geometric means of the following pairs of numbers:

**(i) **2 and 8

**(ii) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).