Step-by-step Solution

1. Define the relationship between the angles.

Let the angles of the triangle be $A, B, C$ and the corresponding opposite sides be $a, b, c$. The triangle is obtuse, so one angle is greater than $\frac{\pi}{2}$. Let the angles be ordered such that $A$ is the smallest and $C$ is the largest.

From the problem statement, the difference between the largest and the smallest angle is $\frac{\pi}{2}$. $C - A = \frac{\pi}{2} \implies C = A + \frac{\pi}{2}$ Since $A > 0$, it is clear that $C > \frac{\pi}{2}$, so the triangle is indeed obtuse and $C$ is the largest angle.

The sum of angles in a triangle is $\pi$: $A + B + C = \pi$ Substituting $C = A + \frac{\pi}{2}$: $A + B + (A + \frac{\pi}{2}) = \pi$ $2A + B + \frac{\pi}{2} = \pi$ $B = \frac{\pi}{2} - 2A$ For $B$ to be a positive angle, $\frac{\pi}{2} - 2A > 0 \implies 2A < \frac{\pi}{2} \implies A < \frac{\pi}{4}$.

2. Use the condition that the sides are in Arithmetic Progression (A.P.).

The sides $a, b, c$ are in A.P. By the Law of Sines, $a = 2R \sin A$, $b = 2R \sin B$, $c = 2R \sin C$, where $R$ is the circumradius. So, $\sin A, \sin B, \sin C$ are also in A.P.

Since $C$ is the largest angle, $c$ is the longest side. For the sides to be in A.P., the middle term must be $a$ or $b$. We assume $A$ is the smallest angle, which implies $A<B$, thus $a<b<c$. This requires $a, b, c$ to be in increasing order of an A.P. This means $2b = a+c$. This translates to: $2\sin B = \sin A + \sin C$ This is a standard result for triangles with sides in A.P. It can be manipulated into a more useful form using half-angle formulas: $2(2\sin(B/2)\cos(B/2)) = 2\sin\left(\frac{A+C}{2}\right)\cos\left(\frac{A-C}{2}\right)$ Since $A+C = \pi - B$, we have $\frac{A+C}{2} = \frac{\pi}{2} - \frac{B}{2}$, so $\sin\left(\frac{A+C}{2}\right) = \cos\left(\frac{B}{2}\right)$. $4\sin(B/2)\cos(B/2) = 2\cos(B/2)\cos\left(\frac{C-A}{2}\right)$ Assuming $\cos(B/2) \neq 0$ (which is true for a triangle), we can divide by $2\cos(B/2)$: $2\sin(B/2) = \cos\left(\frac{C-A}{2}\right)$ We are given $C-A = \frac{\pi}{2}$, so $\frac{C-A}{2} = \frac{\pi}{4}$. $2\sin(B/2) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ $\sin(B/2) = \frac{1}{2\sqrt{2}}$

3. Calculate the inradius $r$.

The formula for the inradius $r$ is given by: $r = 4R \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)$ We are given that the circumradius $R=1$. So, $r = 4 \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)$ Let's group the terms involving $A$ and $C$ and use the product-to-sum formula: $r = 4 \sin\left(\frac{B}{2}\right) \left[ \sin\left(\frac{A}{2}\right) \sin\left(\frac{C}{2}\right) \right]$ $r = 4 \sin\left(\frac{B}{2}\right) \left[ \frac{1}{2} \left( \cos\left(\frac{C-A}{2}\right) - \cos\left(\frac{C+A}{2}\right) \right) \right]$ $r = 2 \sin\left(\frac{B}{2}\right) \left[ \cos\left(\frac{C-A}{2}\right) - \cos\left(\frac{\pi-B}{2}\right) \right]$ $r = 2 \sin\left(\frac{B}{2}\right) \left[ \cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{B}{2}\right) \right]$

Now, substitute the value of $\sin(B/2) = \frac{1}{2\sqrt{2}}$ and $\cos(\pi/4) = \frac{1}{\sqrt{2}}$: $r = 2 \left(\frac{1}{2\sqrt{2}}\right) \left[ \frac{1}{\sqrt{2}} - \frac{1}{2\sqrt{2}} \right]$ $r = \frac{1}{\sqrt{2}} \left[ \frac{2-1}{2\sqrt{2}} \right]$ $r = \frac{1}{\sqrt{2}} \left[ \frac{1}{2\sqrt{2}} \right]$ $r = \frac{1}{4}$

Alternative Method Check: From $2\sin B = \sin A + \sin C$ and angle relations, we get $2\cos(2A) = \sin A + \cos A$. Using $\cos(2A) = \cos^2 A - \sin^2 A$, this becomes $2(\cos A - \sin A)(\cos A + \sin A) = \sin A + \cos A$. As $\sin A + \cos A \neq 0$ for a triangle angle, we get $2(\cos A - \sin A) = 1$, so $\cos A - \sin A = \frac{1}{2}$. A less common formula for the inradius is $r = R(\cos A + \cos B + \cos C - 1)$. Using the angle relations, it can be shown that $r = \frac{1 - (\cos A - \sin A)}{2}$. Substituting $\cos A - \sin A = 1/2$, we get r = \frac{1 - 1/2}{2} = \frac{1/4}. This confirms the result.

The inradius is $r=0.25$.

Final Answer is 0.25.