You are given four springs, one each of 3.5, 6, 8.5, and 11 newtons per meter [n/m]. how do i find the smallest equivalent stiffness that can be made using only three of these springs?

You have two possible ways to connect the springs: in parallel or in series.

The equivalent stiffness of three springs in parallel is given by:
k_eq = k₁ + k₂ + k₃
In order to keep this number the smallest possible, you need to take the three springs with smaller k:
k_eq_min = 3.5 + 6 + 8.5 = 18 N/m

The equivalent stiffness of three springs in series is given by:
1 / k_eq = 1 / k₁ + 1 / k₂ + 1 / k₃
In order to get the smallest k_eq possible, 1 / k_eq must be the biggest possible, therefore you need to take again the three springs with smaller k:

k_eq = 1 / (1 / k₁ + 1 / k₂ + 1 / k₃) 
         = 1 / (1 / 3.5 + 1 / 6 + 1 / 8.5)
         = 1.754 N/m

Therefore, in order to get the smallest equivalent stiffness, you need to connect the first three springs in series (one after the other).

What even is this. where is the question

What formula  this one  (x_/