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Question
Which term of the G.P. :
\[\frac{1}{3}, \frac{1}{9}, \frac{1}{27} . . \text { . is } \frac{1}{19683} ?\]
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Solution
\[\text { Here, first term, }a = \frac{1}{3} \]
\[\text { and common ratio } r = \frac{1}{3}\]
\[\text { Let the } n^{th}\text { term be } \frac{1}{19683} . \]
\[ \therefore a_n = \frac{1}{19683}\]
\[ \Rightarrow a r^{n - 1} = \frac{1}{19683}\]
\[ \Rightarrow \left( \frac{1}{3} \right) \left( \frac{1}{3} \right)^{n - 1} = \frac{1}{19683}\]
\[ \Rightarrow \left( \frac{1}{3} \right)^{n - 1} = \frac{3}{\left( 3 \right)^9} = \left( \frac{1}{3} \right)^8 \]
\[ \Rightarrow n - 1 = 8 \]
\[ \Rightarrow n = 9\]
\[\text { Thus, the } 9^{th} \text { term of the given G . P . is } \frac{1}{19683} .\]
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