Question

The sum of first three terms of a geometric sequence is $\frac{13}{12}$ and their product is $-1$. Find the common ratio and the terms

Answer 1

We may take the first three terms of the geometric sequence as $\frac{a}{r},a,ar$.
Then, $\frac{a}{r}+a+ar=\frac{13}{12}$
$a(\frac{1}{r}+1+r)=\frac{13}{12}\Rightarrow a\left(\frac{r^2+r+1}{r}\right)=\frac{13}{12}....\left(1\right)$
Also,
$\left(\frac{a}{r}\right)\left(a\right)\left(ar\right)=-1$
$\Rightarrow a^3=-1\therefore a=-1$
Substituting $a=-1$ in $\left(1\right)$ we obtain,
$(-1)\left(\frac{r^2+r+1}{r}\right)=\frac{13}{12}$
$\Rightarrow 12r^2+12r+12=-13r$
$12r^2+25r+12=0$
$(3r+4)(4r+3)=0$
Thus, $r=-\frac{4}{3}$ or $-\frac{3}{4}$
When $r=-\frac{4}{3}$ and $a=-1$, the terms are $\frac{3}{4},-1,\frac{4}{3}$.
When $r=-\frac{3}{4}$ and $a=-1$, we get $\frac{4}{3},-1,\frac{3}{4}$, which is in the reverse order.