The sum of first three terms of a $G.P.$ is $\frac{13}{12}$ and their product is $-1$ Find the common ratio and the terms.
The sum of first three terms of a $G.P.$ is $\frac{13}{12}$ and their product is $-1$ Find the common ratio and the terms.
Let $\frac{a}{r}, a,$ ar be the first three terms of the $G.P.$ Then
$\frac{a}{r}+a r+a=\frac{13}{12}$ ........$(1)$
and $\left(\frac{a}{r}\right)(a)(a r)=-1$ ........$(2)$
From $(2),$ we get $a^{3}=-1,$ i.e., $a=-1$ (considering only real roots)
Substituting $a=-1$ in $(1),$ we have
$-\frac{1}{r}-1-r=\frac{13}{12}$ or $12 r^{2}+25 r+12=0$
This is a quadratic in $r$, solving, we get $r=-\frac{3}{4}$ or $-\frac{4}{3}$
Thus, the three terms of $G.P.$ are $: \frac{4}{3},-1, \frac{3}{4}$ for $r=\frac{-3}{4}$ and $\frac{3}{4},-1, \frac{4}{3}$ for $r=\frac{-4}{3}$
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