The sum of first three terms of a G.P. is 13/12 and their product is − 1.

Solution:

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

∴ Product of the G.P. = −1

= a3 = −1

= a = −1

Similarly, Sum of the G.P. $=\frac{13}{12}$

$\Rightarrow \frac{a}{r}+a+a r=\frac{13}{12}$

Substituting the value of a = −1

$\frac{-1}{r}-1-r=\frac{13}{12}$

$\Rightarrow 12 r^{2}+25 r+12=0$

$\Rightarrow 12 r^{2}+16 r+9 r+12=0$

$\Rightarrow 4 r(3 r+4)+3(3 r+4)=0$

$\Rightarrow(4 r+3)(3 r+4)=0$

$\Rightarrow r=-\frac{3}{4},-\frac{4}{3}$

Hence, the G.P. for $a=-1$ and $r=-\frac{3}{4}$ is $\frac{4}{3},-1$ and $\frac{3}{4}$.

And, the G.P. for $a=-1$ and $r=-\frac{4}{3}$ is $\frac{3}{4},-1$ and $\frac{4}{3}$.