Stiffness

k1xk2/k1+k2

k1.k2/k1+k2

It should be noted that when springs are in series then their equivalent stiffeness decreases and when they r in parallel, stiffness increases.so we have to apply a larger force in parallel spring system and lower force in series system.

1/k1+1/k2

1.All Springs in series will experience the same force F applied at the end. (F=F1=F2)

2.Total deflection of this system will be equal to deflection of individual spring (L=L1+L2)

As F=KL

THEREFORE

F1=K1 x L1 , F2=K2 x L2
OR L1=F/K1, L2=F/K2
SO L=L1+L2 WILL BECOME
F/K=F/K1+F/K2
OR 1/K=1/K1+1/K2
OR K=(K1 x K2)/(K1+K2)