A spring of force constant fr is cut into lengths of ratio 1: 2: 3. They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k" . Then k' : k" is :

Length of the spring segments $=\frac{\mathrm{l}}{6},\frac{\mathrm{l}}{3},\frac{\mathrm{l}}{2}$
As we know $\mathrm{k}\propto \frac{1}{\mathrm{l}}$
So, spring constant for spring segments will be 
           ${\mathrm{k}}_1=6\mathrm{k},{\mathrm{k}}_2=3\mathrm{k},{\mathrm{k}}_3=2\mathrm{k}$
So in parallel combination
        $\mathrm{k}"={\mathrm{k}}_1+{\mathrm{k}}_2+{\mathrm{k}}_3=11 \mathrm{k}$
In series combination
       $\mathrm{k}\text{'}=\mathrm{k}$(As it will become original spring)
So,  $\mathrm{k}\text{'}:\mathrm{k}"=1 : 11$