मराठी

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

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प्रश्न

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

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

बेरीज
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उत्तर

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

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

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

\[ \therefore a_n = 729\]

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

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

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

\[ \Rightarrow n - 1 = 11\]

\[ \Rightarrow n = 12\]

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

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पाठ 9: Arithmetic and geometric progression - Exercise 9D [पृष्ठ १९३]

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नूतन Mathematics [English] Class 10 ICSE
पाठ 9 Arithmetic and geometric progression
Exercise 9D | Q 9. | पृष्ठ १९३

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