Question

Insert 6 geometric means between 27 and  \[\frac{1}{81}\] .

Answer 1

\[\text { Let the 6 G . M . s between 27 and } \frac{1}{81}\text {  be} G_1 , G_2 , G_3 , G_4 , G_5 \text { and } G_6 . \]

\[\text { Thus }, 27, G_1 , G_2 , G_3 , G_4 , G_5 , G_6 \text { and } \frac{1}{81} \text { are in G . P } . \]

\[ \therefore a = 27, n = 8 \text { and } a_8 = \frac{1}{81}\]

\[ \because a_8 = \frac{1}{81}\]

\[ \Rightarrow {ar}^7 = \frac{1}{81}\]

\[ \Rightarrow r^7 = \frac{1}{81 \times 27}\]

\[ \Rightarrow r^7 = \left( \frac{1}{3} \right)^7 \]

\[ \Rightarrow r = \frac{1}{3}\]

\[ \therefore G_1 = a_2 = ar = 27\left( \frac{1}{3} \right) = 9\]

\[ G_2 = a_3 = a r^2 = 27 \left( \frac{1}{3} \right)^2 = 3\]

\[ G_3 = a_4 = a r^3 = 27 \left( \frac{1}{3} \right)^3 = 1\]

\[ G_4 = a_5 = a r^4 = 27 \left( \frac{1}{3} \right)^4 = \frac{1}{3}\]

\[ G_5 = a_6 = a r^5 = 27 \left( \frac{1}{3} \right)^5 = \frac{1}{9} \]

\[ G_6 = a_7 = a r^6 = 27 \left( \frac{1}{3} \right)^6 = \frac{1}{27}\]