If the sum of the terms of an infinitely decreasing G.P. is equal to the greatest value of the function $f\left(x\right)=x^3+3x-9$  on $[-5,3]$  and the difference between the first and second term is $f^{\text{'}}\left(0\right)$ then the common ratio of the progression is

$f\left(x\right)=x^3+3x-9$,
$f^{\text{'}}\left(x\right)=3x^2+3$  $f\left(x\right)>0$ 
in $[-5,3]$ then $f\left(x\right)$ is increasing function on $[-5,3]$  
  $\therefore f\left(x\right)$  have greatest value at $x=3$ 
 $f\left(x\right)={\left(3\right)}^3+3\left(3\right)-9=27$
  Let $a,ar,a{r}^2\mathrm{.......}$  be a G.P. with common ratio $\left|r\right|<1$
$S_{\infty }=27\left(given\right)$  
$\Rightarrow$$\frac{a}{1-r}=27\mathrm{............}\left(1\right)$ 
 $a-ar=f^{\text{'}}\left(0\right)$$\Rightarrow$  $a(1-r)=f^{\text{'}}\left(0\right)=3\mathrm{.........}\left(2\right)$  
   $\frac{\left(2\right)}{\left(1\right)}\Rightarrow \frac{a(1-r)}{\frac{a}{(1-r)}}=\frac{3}{27}$
${(1-r)}^2=\frac{1}{9}\Rightarrow 1-r=\pm \frac{1}{3}$
$r=1\pm \frac{1}{3}=\frac{4}{3},\frac{2}{3}$$\left(\begin{array}{l}\therefore \left|r\right|<1\\ r\ne \frac{4}{3}\end{array}\right)$ 
$\therefore r=\frac{2}{3}$