Question

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

Answer 1

Let $G_1,G_2,G_3,G_4,G_5,G_6$ be 6 geometric means between $a=27$ and $b=\frac{1}{81}$

Then, 27, $G_1,G_2,G_3,G_4,G_5,G_6,\frac{1}{81}$ is a G.P.

with common ratio r given by

$r={\left(\frac{b}{a}\right)}^{\frac{1}{n+1}}$

$={\left(\frac{\frac{1}{81}}{27}\right)}^{\frac{1}{6+1}}={\left(\frac{1}{81\times 27}\right)}^{\frac{1}{7}}={\left(\frac{1}{3^7}\right)}^{\frac{1}{7}}$

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

$G_2=a{r}^2=27\times \frac{1}{9}=3$

$G_3=a{r}^3=27\times \frac{1}{27}=1$

$G_4=a{r}^4=27\times \frac{1}{27\times 3}=\frac{1}{3}$

$G_5=a{r}^5=27\times \frac{1}{3^5}=\frac{1}{9}$

$G_6=a{r}^6=27\times \frac{1}{36}=\frac{1}{27}$.

Hence, 9, 3, 1, $\frac{1}{3},\frac{1}{9},\frac{1}{27}$ are 6 geometric means between 27 and $\frac{1}{81}$.