There are $$3$$ positive real numbers. The second is greater than the first by the same amount that the third is greater than the second. The product of the two smaller numbers is $$85$$ and that of the two bigger numbers is $$115$$. Then the difference between the smallest and the greatest numbers is

Let the three numbers be

$$a,\quad a+d,\quad a+2d,$$

where $$d$$ is the common difference.

We are given that

$$a(a+d)=85,$$

and

$$(a+d)(a+2d)=115.$$

Divide the second equation by the first equation.

$$\dfrac{a+2d}{a}=\dfrac{115}{85}=\dfrac{23}{17}.$$

Cross-multiplying,

$$17(a+2d)=23a.$$

$$17a+34d=23a.$$

$$6a=34d.$$

$$3a=17d.$$

So,

$$a=\dfrac{17d}{3}.$$

Substitute this into

$$a(a+d)=85.$$

$$\dfrac{17d}{3}\left(\dfrac{17d}{3}+d\right)=85.$$

$$\dfrac{17d}{3}\times\dfrac{20d}{3}=85.$$

$$\dfrac{340d^2}{9}=85.$$

$$340d^2=765.$$

$$d^2=\dfrac{9}{4}.$$

Since the numbers are positive,

$$d=\dfrac{3}{2}.$$

The difference between the greatest and the smallest numbers is

$$(a+2d)-a=2d=3.$$

Hence, the required difference is $$3$$