If the roots of the equation, $x^3+p{x}^2+qx-1=0$  form an increasing G.P., where $pandq$  are real, then

Let the roots be $a/r,a,ar,$  where $a>0,r>1.$  

Now, $\left(a/r\right)+a+ar=-p$   ------------$\left(1\right)$

$a\left(a/r\right)+a\left(ar\right)+\left(ar\right)\left(a/r\right)=q$ ----------  $\left(2\right)$

$\left(a/r\right)\left(a\right)\left(ar\right)=1$      ---------------$\left(3\right)$

$\Rightarrow a^3=1$

$\Rightarrow a=1$

Hence, $\left(c\right)$  is correct. From $\left(1\right),$  putting $a=1,$  we get

$-p-3>0$                          $\left(\because r+\frac{1}{r}>2\right)$

$\Rightarrow -p>3\Rightarrow p<-3$

Hence, $\left(b\right)$  is not correct. Also,

$\left(1/r\right)+1+r=-p$       ------------- $\left(4\right)$

From $\left(2\right),$  putting $a=1,$  we get

$\left(1/r\right)+r+1=q$         ----------------$\left(5\right)$

From $\left(4\right)and\left(5\right),$ we have $-p=q$ $\Rightarrow p+q=0$

Hence, $\left(a\right)$ is correct.

Now, as$r>1$    $a/r=1/r<1$  and $ar=r>1$