The sum of first three terms of a G.P. is 13 / 12, and their product is -1 . Find the G.P.

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The sum of first three terms of a G.P. is 13 / 12, and their product is -1 . Find the G.P.

Solution:

Let the three numbers be $a/r$ , $a$ , $ar$
According to the question
$\mathrm{a} / \mathrm{r}+\mathrm{a}+\mathrm{ar}=13 / 12 \ldots$ eq. (1)
$\mathrm{a} / \mathrm{r} \times \mathrm{a} \times \mathrm{ar}=-1 \ldots$ eq. $(2)$
From eq.(2) we obtain,
$\begin{array}{l} a^{3}=-1 \\ a=-1 \end{array}$
From eq.(1) we obtain,
$\left(a+a r+a r^{2}\right) / r=13 / 12$ $12 a+12 a r+12 a r^{2}=13 r \ldots$ eq.(3)
On substituting $a=-1$ in eq.(3) we get
$\begin{array}{l} 12(-1)+12(-1) r+12(-1) r^{2}=13 r \\ 12 r^{2}+25 r+12=0 \\ 12 r^{2}+16 r+9 r+12=0 \ldots \text { eq.}(4) \\ 4 r(3 r+4)+3(3 r+4)=0 \\ r=-3 / 4 \text { or } r=-4 / 3 \end{array}$
The equation will now be
$\begin{array}{l} -1 /(-3 / 4),-1,-1 \times-3 / 4 \text { or }-1 /(-4 / 3),-1,-1 \times-4 / 3 \\ 4 / 3,-1,3 / 4 \text { or } 3 / 4,-1,4 / 3 \end{array}$
As a result, the three numbers are $4 / 3,-1,3 / 4$ or $3 / 4,-1,4 / 3$