Answer
With $a=3r$, $b=4r$, $c=5r$ we have a right triangle that is similar to a 3-4-5 triangle for any $r\gt0$.
Work Step by Step
Let $a,b,c$ be the sides of a right triangle such that $a,b,c$ are in arithmetic progression. That is:
$b=a+r$
$c=b+r=a+2r$
Assuming that $r\gt0$, $c$ is the hypotenuse because $c\gt b\gt a$. So:
$c^2=a^2+b^2$
$(a+2r)^2=a^2+(a+r)^2$
$a^2+4ar+4r^2=a^2+a^2+2ar+r^2$
$0=a^2-2ar-3r^2$
$0=a^2-3ar+ar-3r^2$
$0=a(a-3r)+r(a-3r)$
$0=(a+r)(a-3r)$
$a+r=0$
$a=-r$
It is not a valid solution because both $a$ and $r$ must be positive numbers.
$a-3r=0$
$a=3r$
$b=3r+r=4r$
$c=3r+2r=5r$
