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Question
Insert 6 geometric means between 27 and \[\frac{1}{81}\] .
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Solution
\[\text { Let the 6 G . M . s between 27 and } \frac{1}{81}\text { be} G_1 , G_2 , G_3 , G_4 , G_5 \text { and } G_6 . \]
\[\text { Thus }, 27, G_1 , G_2 , G_3 , G_4 , G_5 , G_6 \text { and } \frac{1}{81} \text { are in G . P } . \]
\[ \therefore a = 27, n = 8 \text { and } a_8 = \frac{1}{81}\]
\[ \because a_8 = \frac{1}{81}\]
\[ \Rightarrow {ar}^7 = \frac{1}{81}\]
\[ \Rightarrow r^7 = \frac{1}{81 \times 27}\]
\[ \Rightarrow r^7 = \left( \frac{1}{3} \right)^7 \]
\[ \Rightarrow r = \frac{1}{3}\]
\[ \therefore G_1 = a_2 = ar = 27\left( \frac{1}{3} \right) = 9\]
\[ G_2 = a_3 = a r^2 = 27 \left( \frac{1}{3} \right)^2 = 3\]
\[ G_3 = a_4 = a r^3 = 27 \left( \frac{1}{3} \right)^3 = 1\]
\[ G_4 = a_5 = a r^4 = 27 \left( \frac{1}{3} \right)^4 = \frac{1}{3}\]
\[ G_5 = a_6 = a r^5 = 27 \left( \frac{1}{3} \right)^5 = \frac{1}{9} \]
\[ G_6 = a_7 = a r^6 = 27 \left( \frac{1}{3} \right)^6 = \frac{1}{27}\]
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