Question

Insert five geometric means between $486$ and $2/3$.

Answer 1

$486,x_1,x_2,x_3,x_4,x_5,\frac{2}{3}$ will be a $G.P$ of $7$ terms
$\therefore \frac{2}{3}=T_7=a{r}^6=\left(486\right)r^6$.
$\therefore r^6=\frac{1}{729}=\frac{1}{3^6}$ $\therefore r=\pm \frac{1}{3}$, taker $=\frac{1}{3}$
$\therefore x_1=T_2=ar=486.\left(\frac{1}{3}\right)=162$
$x_2=T_3={ar}^2=486\left(\frac{1}{9}\right)=54$,
$x_3=18$, $x_4=6$, $x_5=2$
Hence the five geometric means are $162,54,18,6,2$.