Imagine we have a spring that is hanging from the roof with $k = 200$ and a stone with $ W=10N $ is pulling it down. Then due to $$ F=kx $$ we have $x=.05m=5cm$
Now we halve the length of the spring. Again a stone with the same weight as the last stone, is pulling it down. In this case, what is $x$ ? Does it change by halving the length of the spring?
Indeed, spring constant gets doubled as said by N.S.John.
Suppose the displacement is $x_\text{large}$ in the larger spring; this implies each smaller spring gets displaced by $x_\text{small}$ & it is given by $x_\text{small}= \dfrac{x_\text{large}}{n}$ where $n$ indicates the number of small springs.
Now, each spring under stress has potential energy gained $k_\text{small}x_\text{small}^2 /2\;.$ Equating with that of the large spring, we get: $$nk_\text{small}x_\text{small}^2 /2= k_\text{large}x^2_\text{large}/2\;.$$ From this, it is a trivial algebra to deduce $k_\text{small}= \dfrac{k_\text{large}}{1/n}\;.$