Solution. Let the difference of the arithmetic sequence formed by the sides of the triangle be $d$, and the sides of the triangle be: $BC=a$, $AC=a-d$, and $AB=a+d$, where $d>0$.

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According to the Pythagorean theorem:

$$\begin{aligned} (a-d)^{2}+a^{2} & =(a+d)^{2} \\ a^{2}-2 a d+d^{2}+a^{2} & =a^{2}+2 a d+d^{2} \\ a^{2}-4 a d & =0 \\ a(a-4 d) & =0 \end{aligned}$$

This is only possible if $a=0$ or if $a-4 d=0$.

$a$ cannot be 0, as it is the length of one side of the triangle; thus, $a=4 d$. Therefore, the lengths of the sides are: $BC=a=4 d$, $AC=a-d=3 d$, and $AB=a+d=5 d$.

Thus, the ratio of the sides is: $AC: BC: AB=3: 4: 5$.

Let the radius of the inscribed circle be $r$, and $\frac{a+b+c}{2}=s$. We can write the area of the triangle in two ways:

$$t=r \cdot s=r \cdot \frac{3 d+4 d+5 d}{2}=6 r d, \quad \text { and } \quad t=\frac{3 d \cdot 4 d}{2}=6 d^{2}$$

Since $6 r d=6 d^{2}$, it follows that $r=d$.

This proves that the radius of the inscribed circle is equal to the difference of the arithmetic sequence, and the ratio of the sides is $3: 4: 5$.