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 $h$
$$ L = b + 2 \sqrt{\left(\frac{b}{2}\right)^2 + h^2} $$
