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Question
Insert 5 geometric means between \[\frac{32}{9}\text{and}\frac{81}{2}\] .
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Solution
\[\text{Let the 5 G . M . s between } \frac{32}{9} \text{ and } \frac{81}{2} \text{be }G_1 , G_2 , G_3 , G_4 \text{and} G_5 . \]
\[\frac{32}{9}, G_1 , G_2 , G_3 , G_4 , G_5 , \frac{81}{2}\]
\[ \Rightarrow a = \frac{32}{9}, n = 7 \text { and } a_7 = \frac{81}{2}\]
\[ \because a_7 = \frac{81}{2}\]
\[ \Rightarrow a r^6 = \frac{81}{2}\]
\[ \Rightarrow r^6 = \frac{81}{2} \times \frac{9}{32}\]
\[ \Rightarrow r^6 = \left( \frac{3}{2} \right)^6 \]
\[ \Rightarrow r = \frac{3}{2}\]
\[ \therefore G_1 = a_2 = ar = \frac{32}{9}\left( \frac{3}{2} \right) = \frac{16}{3}\]
\[ G_2 = a_3 = a r^2 = \frac{32}{9} \left( \frac{3}{2} \right)^2 = 8\]
\[ G_3 = a_4 = a r^3 = \frac{32}{9} \left( \frac{3}{2} \right)^3 = 12\]
\[ G_4 = a_5 = a r^4 = \frac{32}{9} \left( \frac{3}{2} \right)^4 = 18\]
\[ G_5 = a_6 = a r^5 = \frac{32}{9} \left( \frac{3}{2} \right)^5 = 27\]
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