Then the inradius of the triangle ABC is

Let the side lengths opposite the angles $$A,B,C$$ (with $$A\lt B\lt C$$) be $$a,b,c$$ respectively. They are in an arithmetic progression, so $$a+b=c+b-a\;\Rightarrow\;b-a=c-b.$$

Because the vertices lie on a circle of radius $$R=1$$, the sine rule gives $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R=2.$$ Hence $$a=2\sin A,\; b=2\sin B,\; c=2\sin C\quad -(1).$$

The largest angle exceeds the smallest by $$\dfrac{\pi}{2}$$, so $$C-A=\frac{\pi}{2}\;\Longrightarrow\;C=A+\frac{\pi}{2}\quad -(2).$$

Using the A.P. condition with $$(1)$$, $$\sin B-\sin A=\sin C-\sin B\;\Longrightarrow\;2\sin B=\sin A+\sin C\quad -(3).$$

Substitute $$C=A+\dfrac{\pi}{2}$$ in $$(3)$$. Because $$\sin\!\left(A+\dfrac{\pi}{2}\right)=\cos A$$ and $$\sin B=\sin\!\left(\dfrac{\pi}{2}-2A\right)=\cos 2A$$, equation $$(3)$$ becomes $$\cos 2A=\frac{\sin A+\cos A}{2}\quad -(4).$$

Write $$\sin A=s,\; \cos A=c$$ so that $$s^2+c^2=1$$. From $$(4)$$, $$c^2-s^2=\dfrac{s+c}{2}$$. Re-arranging: $$2(c-s)(c+s)=s+c\;\Rightarrow\;2(c-s)-1=0\;(\text{since }s+c\neq0).$$ Thus $$c-s=\frac12\quad -(5).$$

With $$c^2+s^2=1$$ and $$(5)$$, $$cs=\frac12\!\Bigl(1-(c-s)^2\Bigr)=\frac12\!\left(1-\frac14\right)=\frac38.$$ Solving the pair $$\begin{cases}c-s=\dfrac12\\[4pt]cs=\dfrac38\end{cases}$$ gives the acute solution $$\sin A=\dfrac{\sqrt7-1}{4},\qquad \cos A=\dfrac{\sqrt7+1}{4}.$$

Consequently $$\sin B=\cos 2A=c^2-s^2=\frac{\sqrt7}{4},\qquad \sin C=\cos A=\frac{\sqrt7+1}{4}.$$

Applying $$(1)$$, the sides are $$a=2\sin A=\frac{\sqrt7-1}{2},\; b=2\sin B=\frac{\sqrt7}{2},\; c=2\sin C=\frac{\sqrt7+1}{2},$$ verifying the arithmetic progression.

Using the formula $$\Delta=\dfrac{abc}{4R}$$ with $$R=1$$, $$\Delta=\frac{1}{4}\left(\frac{\sqrt7-1}{2}\right)\!\left(\frac{\sqrt7}{2}\right)\!\left(\frac{\sqrt7+1}{2}\right) =\frac{3\sqrt7}{16}.$$

Semiperimeter $$s=\frac{a+b+c}{2} =\frac{\tfrac{\sqrt7-1}{2}+\tfrac{\sqrt7}{2}+\tfrac{\sqrt7+1}{2}}{2} =\frac{3\sqrt7}{4}.$$

The inradius is $$r=\frac{\Delta}{s} =\frac{\tfrac{3\sqrt7}{16}}{\tfrac{3\sqrt7}{4}} =\frac14=0.25.$$

Therefore, the required inradius is 0.25.