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Question
Which term of the progression 18, −12, 8, ... is \[\frac{512}{729}\] ?
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Solution
\[\text { Here, first term }, a = 18 \]
\[\text { and common ratio }, r = \frac{- 2}{3}\]
\[\text { Let the } n^{th} \text { term be } \frac{512}{729} . \]
\[ \therefore a r^{n - 1} = \frac{512}{729}\]
\[ \Rightarrow \left( 18 \right) \left( \frac{- 2}{3} \right)^{n - 1} = \frac{512}{729}\]
\[ \Rightarrow \left( \frac{- 2}{3} \right)^{n - 1} = \frac{512}{729} \times \frac{1}{18} = \frac{256}{6561}\]
\[ \Rightarrow \left( \frac{- 2}{3} \right)^{n - 1} = \left( \frac{- 2}{3} \right)^8 \]
\[ \Rightarrow n - 1 = 8 \]
\[ \Rightarrow n = 9\]
\[\text { Thus, the } 9^{th} \text { term of the given G . P . is } \frac{512}{729} .\]
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