With two pegs, there are two strips of rubber working in parallel contributing to a total stiffness $K_{\rm total} = 57.6 \;\rm lbf/in$. So each strip is $K = 28.8\;\rm lbf/in$.
With the three pegs you now have two strips at 26° apart, or 13° from vertical each. The effective spring constant in the vertical direction is thus $K_{eff} =2 K \cos^2(13^\circ) = 54.68\; \rm lbf/in$. One of the $\cos(13^\circ)$ comes from the force projection to vertical and the second from the displacement projection to vertical from along the side of the triangle.
Another way to get the same result if the base is $b$ and the height is $h$ is
$$ K_{eff} = \frac{8 h^2}{b^2+4 h^2} K $$
$$ = \frac{8*52^2}{24^2+4*52^2} 28.8 = 54.68 $$
But also the stretch force value has changed. To get this you need the free length of the band which is not given and cannot be calculated from the given values.
Edit 1
With the values given I came up with the following force on the top pin
$$ F = \frac{288 h \left( \sqrt{b^2+4 h^2}+b-82\right)}{5 \sqrt{b^2+4 h^2}} $$
So with two pins $b=0$, $h=52$ the force is $F=633.6\;\rm lbf$ with stiffness $K_{eff}=\frac{\partial F}{\partial h}=57.6$.
With three pins and $b=24$, $h=52$ the force is $F=1367.6\;\rm lbf$ and stiffness $K_{eff}=\frac{\partial F}{\partial h}=56.02$.
Why ?
The force on the pin is equal to two times the tension projected vertically (from a Free Body Diagram on the pin).
$$ F = 2 T \cos\left(\frac{\theta}{2}\right) $$
with $\cos\left(\frac{\theta}{2}\right) = \frac{h}{\sqrt{\left(\frac{b}{2}\right)^2+h^2}}$ and tension $T = K (L-L_0) $. The initial band length is $L_0 = 2\cdot 41 = 82\;\rm in$ and the stiffness of each strip is actually $K = 14.4\;\rm lbf/in$. This comes from $K_{eff}=4\,K$ when $b=0$ and with $K_{eff}=57.6\;\rm lbf/in$.
The length of the rubber band is the circumference of a triangle with base $b$ and height $h$
$$ L = b + 2 \sqrt{\left(\frac{b}{2}\right)^2 + h^2} $$
You asked for intuitive sense and I'll try to provide it. The formula is:
$$\Delta S = \frac{\Delta Q}{T}$$
So, you can have $\Delta S_1=\frac{\Delta Q}{T_{lower}}$ and $\Delta S_2=\frac{\Delta Q}{T_{higher}}$
Assume the $\Delta Q$ is the same in each case. The denominator controls the "largeness" of the $\Delta S$.
Therefore, $\Delta S_1 > \Delta S_2$
In each case, let's say you had X number of hydrogen atoms in each container. The only difference was the temperature of the atoms. The lower temperature group is at a less frenzied state than the higher temperature group. If you increase the "frenziedness" of each group by the same amount, the less frenzied group will notice the difference more easily than the more frenzied group.
Turning a calm crowd into a riot will be much more noticeable than turning a riot into a more frenzied riot. Try to think of the change in entropy as the noticeably of changes in riotous behavior.
Best Answer
It sounds like you're describing two separate effects:
Elastomers can have a negative thermal expansion coefficient, i.e., $$\alpha=\frac{1}{L}\left(\frac{\partial L}{\partial T}\right)_F<0,$$ because the higher-entropy state (corresponding to a higher temperature) is one in which the polymer chains move around more, which has the consequence of shortening their end-to-end length. When characterizing this material property, the force $F$ is kept constant.
Most materials (including elastomers) generally get more compliant when they are heated; thus, their stiffness $$k=\left(\frac{\partial F}{\partial L}\right)_T,$$ which is characterized at a constant temperature, decreases with increasing temperature:
In polymers, the origin is the melting of weak bonds between the chains. However, exceptions can exist; for example, the stiffness of a purely entropic system increases with temperature, so once melting is complete, we might see an increase in the elastic modulus with heating:
(Hosford, Mechanical Behavior of Materials)
Therefore, if you hang a weight on a rubber band (to achieve a change in length of $\Delta L$) and heat it up, the weight will probably rise because of (1). If you then remove the weight at the same higher temperature, the rubber band may contract by more or less than $\Delta L$ as mediated by the mechanisms discussed in (2). If you then cool the system down to the original temperature, the rubber band will return to its original length, $\Delta L$ above the original position of the weight, if it is not already there and if the entire loading time was short enough that irreversible flow was negligible.
Put another way, since you're asking about elasticity in the context of a hot and a cold rubber band loaded by the same weight, I should emphasize that one can't directly measure a system's stiffness by keeping the force constant and observing the displacement when changing other things. One measures the stiffness by changing the force while watching the displacement (or vice versa) while keeping other things constant.