The sum of first three terms of a GP is

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

It is given that $\frac{a}{r} \times a \times a r=1$

$\Rightarrow a^{3}=1$

$\Rightarrow a=1$

And

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

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

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

$\Rightarrow\left(\frac{1}{r}+r\right)=\frac{39}{10}-1=\frac{29}{10}$

$\Rightarrow 10\left(1+r^{2}\right)=29 r$

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

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

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

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

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

Therefore the first three terms are:

i) If $r=\frac{5}{2}$ then

$\frac{2}{5}, 1, \frac{5}{2}$

ii) If $r=\frac{2}{5}$ then

$\frac{5}{2}, 1, \frac{2}{5}$

Ans: Common ratio $r=\frac{5}{2}, \frac{2}{5}$ and the first three terms are:

i) if $r=\frac{5}{2}$ then

$\frac{2}{5}, 1, \frac{5}{2}$

ii) If $r=\frac{2}{5}$ then

$\frac{5}{2}, 1, \frac{2}{5}$