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).

2. they have negligible mass.

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