This isn't a geometric progression as well?

Excellent problem!!!

Given, $x^2,x^3,x^4$ are in A.P.,

We know that if $3$ numbers $a,b,c$ are in A.P., then $a+c=2b$

For now, we consider the two cases $a,b,c$ and $c,b,a$ the same, as we are just adding $a$ and $c$ . We will deal with it later.

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We get three cases:

Note for below working : We can divide by $x^2$ freely as we know that $x$ is not an integer, so it can't be $0$ .

Case 1

$x^2+x^4=2x^3$ $\implies 1+x^2=2x$ $\implies (x-1)^2=0 \implies x=1 \color{#D61F06}{\text{ - Contradiction! x can't be an integer!}}$

Case 2

$x^3+x^4=2x^2$ $\implies x+x^2=2$ $\implies (x-1)(x+2)=0 \implies x=1, -2 \color{#D61F06}{\text{ - Contradiction! x can't be an integer!}}$

Case 3

$x^3+x^2=2x^4$ $\implies 1+x=2x^2$ $\implies (x-1)(2x+1)=0 \implies x=1,\dfrac{-1}{2} ,\color{#D61F06}{\text{ x=1 - Contradiction! x can't be an integer!}}$ $\implies \boxed{x=\dfrac{-1}{2}}$

So, the only value of $x$ is $\dfrac{-1}{2}$ .

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We can now fond the first three terms: $x^2 =\dfrac{1}{4}, x^3 =\dfrac{-1}{8},x^4 =\dfrac{1}{16}$

Now, since they are in A.P., the negative term can't be the second one(obvious, I won't prove).

So, we get two A.P.s to consider in the end, as the other two don't follow $a+c=2b$ and the other two contain the negative term as second.

A.P. 1

$\dfrac{-1}{8},\dfrac{1}{16},\dfrac{1}{4}$ Here, the common difference is $\dfrac{3}{16}$ So, we get an increasing progression. Hence, only a positive term, that is even power of $x$ can come later. However, $x^6=\dfrac{1}{64}$ , which can't be reached since we already have $\dfrac{1}{4}$ , which is greater than $x^6=\dfrac{1}{64} \color{#D61F06}{\text{ - Contradiction!}}$

A.P. 2

$\dfrac{1}{4},\dfrac{1}{16},\dfrac{-1}{8}$ Here, the common difference is $\dfrac{-3}{16}$ . So, we get an decreasing progression. So, we can find out some of the next terms: $\dfrac{1}{4},\dfrac{1}{16},\dfrac{-1}{8},\dfrac{-5}{16},\boxed{\dfrac{-1}{2}}$

Hence, we get $x$ later in the progression.

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P.S. I really liked the problem. If it is original, I really appreciate you, and I want more of this kind!