The sum of first three terms of a G.P.

Solution:

Let the terms of the G.P be $\frac{a}{r}, a$ and $a r$.

$\therefore$ Product of the G.P. $=1$

$\Rightarrow a^{3}=1$

$\Rightarrow a=1$

Now, sum of the G.P. $=\frac{39}{10}$

$\Rightarrow \frac{a}{r}+a+a r=\frac{39}{10}$

$\Rightarrow a\left(\frac{1}{r}+1+r\right)=\frac{39}{10}$

$\Rightarrow 1\left(\frac{1}{r}+1+r\right)=\frac{39}{10}$

$\Rightarrow 10 r^{2}+10 r+10=39 r$

$\Rightarrow 10 r^{2}-29 r+10=0$

$\Rightarrow 10 r^{2}-25 r-4 r+10=0$

$\Rightarrow 5 r(2 r-5)-2(2 r-5)=0$

$\Rightarrow(5 r-2)(2 r-5)=0$

$\Rightarrow r=\frac{2}{5}, \frac{5}{2}$

Hence, putting the values of $a$ and $r$, the required numbers are $\frac{5}{2}, 1, \frac{2}{5}$ or $\frac{2}{5}, 1$ and $\frac{5}{2}$.