A rational numbers cannot have irrational as$\exists n \in \mathbb{Z}, \, \sqrt[n]{\frac xy}$, but the two equalities give: $b^2=\frac{(a^2+c^2)}{2} \implies b = \sqrt[2]{\frac {(a^2+c^2)}{2}}$.
To avoid this, need $4\mid a$, & $a=c$; so that if $\exists t=a/2, \, b = \sqrt[2]{\frac {(a^2+c^2)}{2}} => 2t$.
As an example, $a=c=8, t=4, b=8$.
I am giving an idea for a particular case only, & am unable to pursue.
On
The general solution of $b^2-a^2=5$ has $$b=\frac12\left(t+\frac 5t\right)$$ for $t\in\Bbb Q^*$. Then $$c^2=b^2+5=\frac{t^4+10t^2+25}{4t^2}+5=\frac{t^4+30t^2+25}{4t^2}.$$ The problem boils down to whether the genus-one curve $$y^2=x^4+30x^2+25$$ has rational points.
On
We can rewrite these numbers with common denominator $D$:
$a=\dfrac{A}{D}$, $\;b=\dfrac{B}{D}$, $\;c=\dfrac{C}{D}$. Then we'll get diophantine equation (system of diophantine equations):
$$
B^2-A^2=C^2-B^2=5D^2.\tag{1}
$$
The smallest solution of $(1)$ is $(A,B,C,D)=(31,41,49,12)$.
So, example of such three rational numbers $a,b,c$ is: $$a=\dfrac{31}{12},\quad b=\dfrac{41}{12}, \quad c=\dfrac{49}{12}.$$
Short algorithm (to find $A,B,C,D$) description:
Algebraic approach.
First, note that $$a^2+c^2=2b^2,$$ and if denote $$p=\dfrac{a}{b}, \quad q=\dfrac{c}{b}, \tag{1}$$ then we will search rational points on the circle $$p^2+q^2=2.\tag{2}$$
According to the article Keith Conrad "Arithmetic progressions of three squares", each rational solution of eq. $(2)$ (by exception of $(p,q)=(1,-1)$) can be written in the parametric form $$ p = \dfrac{m^2-2m-1}{m^2+1}, \qquad q=\dfrac{-m^2-2m+1}{m^2+1},\\ \left( m = \dfrac{q-1}{p-1}\right),\tag{3} $$ where $m$ is rational parameter (positive or negative);
and vice versa (for each rational $m$ pair $(p,q)$ will have rational coordinates).
If we require distance (step) $n=5$ between numbers $a^2,b^2,c^2$ in algebraic progression, then (according to Corollary $3.3$ of the article above) each rational solution of equation $$b^2-a^2=c^2-b^2=n\tag{4}$$ can be written in the form $$ a=\dfrac{x^2-2nx-n^2}{2y}, \quad b=\dfrac{x^2+n^2}{2y}, \quad c=\dfrac{x^2+2nx-n^2}{2y}, \\ \left(\; x=\dfrac{n(c-b)}{a-b}, \qquad y=\dfrac{n^2(2b-a-c)}{(a-b)^2} \right), \tag{5} $$
where $(x,y)$ is rational point on the elliptic curve $$ y^2=x^3-nx;\tag{6} $$ and vice versa (each rational point of the EC $(6)$ detemines rational solution of eq. $(4)$).
As was discussed before, one simple positive rational solution is $$(a,b,c) = \left(\dfrac{31}{12}, \dfrac{41}{12}, \dfrac{49}{12}\right);$$
in fact, it means that there are $8$ rational solutions: $$(a,b,c)=\left(\pm\dfrac{31}{12}, \pm\dfrac{41}{12}, \pm\dfrac{49}{12}\right),$$ for which (see $(5), (6)$) we have the set of $8$ rational points on the EC $(6)$: $$ (x,y) = \left(-4,\pm 6\right), \\ (x,y) = \left( 45, \pm 300\right), \\ (x,y) = \left(-\frac{5}{9},\pm \frac{100}{27}\right), \\ (x,y) = \left( \frac{25}{4}, \pm \frac{75}{8}\right). $$
Applying the beauty of the group law of EC, we can derive other rational points from existing ones.
Example:
if we'll take starting point $P_1=(-4,6)$, then denote point $P_2$ as $P_2 = P_1+P_1$
(note that we 'add' two points not coordinate-wise, but according to the group law of EC):
$\; P_2 = \left( \frac{1681}{144} , -\frac{62279}{1728} \right)$, which leads to the solution $$(a,b,c) = \left(\frac{113279}{1494696}, - \frac{3344161}{1494696}, \frac{4728001}{1494696}\right);$$ then let's denote $P_3$ as $P_3 = P_1+P_2$: $\; P_3 = \left(-\frac{2439844}{5094049} , \frac{39601568754}{11497268593} \right)$, which leads to the solution $$(a,b,c) = \left( -\frac{518493692732129}{178761481355556}, \frac{654686219104361}{178761481355556}, \frac{767067390499249}{178761481355556} \right). $$
This way, if focus on positive solutions $(a,b,c)$ of eq. $(4)$ in the form $(a,b,c)=\left(\frac{A}{D},\frac{B}{D},\frac{C}{D}\right)$, where $A,B,C,D\in\mathbb{N}$, we can construct series of rational points (with increasing size of common denominator):
\begin{array}{|l|} \hline A=31; \\ B=41; \\ C=49; \\ D=12; \\ \hline A=113279; \\ B=3344161; \\ C=4728001; \\ D=1494696; \\ \hline A=518493692732129; \\ B=654686219104361; \\ C=767067390499249; \\ D=178761481355556; \\ \hline A=249563579992463717493803519; \\ B=249850594047271558364480641; \\ C=250137278774864229623059201; \\ D=5354229862821602092291248; \\ \hline A=115038188620995226180802686473825513089249; \\ B=160443526614433014168714029147613242401001; \\ C=195577262542844878506138849501555847171249; \\ D=50016678000996026579336936742637753055940; \\ \hline A=21214405287844054428542609853501469645112322962237848990081;\\ B=209239116668342644167838867143329714389679018137228536721441; \\ C=295147361324101461665473218814630755582253386512598845803201; \\ D=93092380947563478644577555596900542802151091304399908363272; \\ \hline \ldots \end{array}