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Question
Which term of the G.P. :
\[\sqrt{3}, 3, 3\sqrt{3}, . . . \text { is } 729 ?\]
Solution
\[\text { Here, first term, }a = \sqrt{3} \]
\[\text { and common ratio }, r = \sqrt{3}\]
\[\text { Let the } n^{th} \text { term be } 729 . \]
\[ \therefore a_n = 729\]
\[ \Rightarrow a r^{n - 1} = 729\]
\[ \Rightarrow \left( \sqrt{3} \right) \left( \sqrt{3} \right)^{n - 1} = 729\]
\[ \Rightarrow (\sqrt{3} )^{n - 1} = \frac{\left( \sqrt{3} \right)^{12}}{\sqrt{3}} = (\sqrt{3} )^{11} \]
\[ \Rightarrow n - 1 = 11\]
\[ \Rightarrow n = 12\]
\[\text { Thus, the }{12}^{th}\text { term of the given G . P . is } 729 .\]
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