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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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