Question

Which term of the G.P.: `sqrt3, 3, 3sqrt3`, ... is 729?

Which term of the progression `sqrt3, 3, 3sqrt3`, ... is 729?

Answer 1

\[\text { Here, first term, }a = \sqrt{3} \]

\[\text { and common ratio }, r = \sqrt{3}\]

\[\text { Let the } n^{th} \text { term be } 729 . \]

\[ \therefore a_n = 729\]

\[ \Rightarrow a r^{n - 1} = 729\]

\[ \Rightarrow \left( \sqrt{3} \right) \left( \sqrt{3} \right)^{n - 1} = 729\]

\[ \Rightarrow (\sqrt{3} )^{n - 1} = \frac{\left( \sqrt{3} \right)^{12}}{\sqrt{3}} = (\sqrt{3} )^{11} \]

\[ \Rightarrow n - 1 = 11\]

\[ \Rightarrow n = 12\]

\[\text { Thus, the }{12}^{th}\text {  term of the given G . P . is } 729 .\]