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Show that the Sequence <An>, Defined by an = 2 3 N , N ϵ N is a G.P.

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Question

Show that the sequence <an>, defined by an = \[\frac{2}{3^n}\], n ϵ N is a G.P.

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Solution

We have,

\[ a_n = \frac{2}{3^n}, n \in N\]

\[\text { Putting n } = 1, 2, 3, . . . \]

\[ a_1 = \frac{2}{3^1} = \frac{2}{3}, a_2 = \frac{2}{3^2} = \frac{2}{9}, a_3 = \frac{2}{3^3} = \frac{2}{27} \text { and so on } . \]

\[\text { Now, } \frac{a_2}{a_1} = \frac{\frac{2}{9}}{\frac{2}{3}} = \frac{1}{3}, \frac{a_3}{a_2} = \frac{\frac{2}{27}}{\frac{2}{9}} = \frac{1}{3} \text { and so on } . \]

\[ \therefore \frac{a_2}{a_1} = \frac{a_3}{a_2} = . . . = \frac{1}{3}\]

\[\text { So, the sequence is an G . P . , where } \frac{2}{3} \text { is the first term and } \frac{1}{3}\text {  is the common ratio }.\]

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Chapter 20: Geometric Progression - Exercise 20.1 [Page 10]

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R.D. Sharma Mathematics [English] Class 11
Chapter 20 Geometric Progression
Exercise 20.1 | Q 2 | Page 10

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