Insert 6 geometric means between 27 and
Question:

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

Solution:

Let the 6 G.M.s between 27 and $\frac{1}{81}$ be $\mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, \mathrm{G}_{4}, \mathrm{G}_{5}$ and $\mathrm{G}_{6}$.

Thus, $27, \mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, \mathrm{G}_{4}, \mathrm{G}_{5}, \mathrm{G}_{6}$ and $\frac{1}{81}$ are in G.P.

$\therefore a=27, n=8$ and $a_{8}=\frac{1}{81}$

$\because a_{8}=\frac{1}{81}$

$\Rightarrow \mathrm{ar}^{7}=\frac{1}{81}$

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

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

$\Rightarrow r=\frac{1}{3}$

$\therefore \mathrm{G}_{1}=a_{2}=a r=27\left(\frac{1}{3}\right)=9$

$\mathrm{G}_{2}=a_{3}=a r^{2}=27\left(\frac{1}{3}\right)^{2}=3$

$\mathrm{G}_{3}=a_{4}=a r^{3}=27\left(\frac{1}{3}\right)^{3}=1$

$\mathrm{G}_{4}=a_{5}=a r^{4}=27\left(\frac{1}{3}\right)^{4}=\frac{1}{3}$

$\mathrm{G}_{5}=a_{6}=a r^{5}=27\left(\frac{1}{3}\right)^{5}=\frac{1}{9}$

$\mathrm{G}_{6}=a_{7}=a r^{6}=27\left(\frac{1}{3}\right)^{6}=\frac{1}{27}$

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