Find three numbers in G.P. whose product is 729 and the sum of their products in pairs is 819.

Solution:

Let the required numbers be $\frac{a}{r}, a$ and $a r$.

Product of the G.P. = 729

$\Rightarrow a^{3}=729$

$\Rightarrow a=9$

Sum of the products in pairs = 819

$\Rightarrow \frac{a}{r} \times a+a \times a r+a r \times \frac{a}{r}=819$

$\Rightarrow a^{2}\left(\frac{1}{r}+r+1\right)=819$

$\Rightarrow 81\left(\frac{1+r^{2}+r}{r}\right)=819$

$\Rightarrow 9\left(r^{2}+r+1\right)=91 r$

$\Rightarrow 9 r^{2}-82 r+9=0$

$\Rightarrow 9 r^{2}-81 r-r+9=0$

$\Rightarrow(9 r-1)(r-9)=0$

$\Rightarrow r=\frac{1}{9}, 9$

Hence, putting the values of $a$ and $r$, we get the numbers to be 81,9 and 1 or 1,9 and 81.