Insert 5 geometric means between

Question:

Insert 5 geometric means between $\frac{32}{9}$ and $\frac{81}{2}$.

Solution:

Let the $5 G .$ M.s between $\frac{32}{9}$ and $\frac{81}{2}$ be $\mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, \mathrm{G}_{4}$ and $\mathrm{G}_{5}$.

$\frac{32}{9}, \mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, \mathrm{G}_{4}, \mathrm{G}_{5}, \frac{81}{2}$

$\Rightarrow a=\frac{32}{9}, n=7$ and $a_{7}=\frac{81}{2}$

$\because a_{7}=\frac{81}{2}$

$\Rightarrow a r^{6}=\frac{81}{2}$

$\Rightarrow r^{6}=\frac{81}{2} \times \frac{9}{32}$

$\Rightarrow r^{6}=\left(\frac{3}{2}\right)^{6}$

$\Rightarrow r=\frac{3}{2}$

$\therefore \mathrm{G}_{1}=a_{2}=a r=\frac{32}{9}\left(\frac{3}{2}\right)=\frac{16}{3}$

$\mathrm{G}_{2}=a_{3}=a r^{2}=\frac{32}{9}\left(\frac{3}{2}\right)^{2}=8$

$\mathrm{G}_{3}=a_{4}=a r^{3}=\frac{32}{9}\left(\frac{3}{2}\right)^{3}=12$

$\mathrm{G}_{4}=a_{5}=a r^{4}=\frac{32}{9}\left(\frac{3}{2}\right)^{4}=18$

$\mathrm{G}_{5}=a_{6}=a r^{5}=\frac{32}{9}\left(\frac{3}{2}\right)^{5}=27$

 

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