This isn't a geometric progression?

To find a value of $x$ that would allow such a sequence to exist, we can set up several systems of equations based on all possible orderings of the first 3 terms:

$\text{Case 1: } x^2 + d = x^3, \enspace x^2 + 2d = x^4$ where $d$ is the common difference. This system represents the ordering $x^2, x^3, x^4$ when $d$ is positive and the ordering $x^4, x^3, x^2$ when $d$ is negative.

$\text{Case 2: } x^2 + d = x^3, \enspace x^2 - d = x^4$ which represents the ordering $x^4, x^2, x^3$ when $d$ is positive and the ordering $x^3, x^2, x^4$ when $d$ is negative.

$\text{Case 3: } x^2 + 2d = x^3, \enspace x^2 + d = x^4$ which represents the ordering $x^2, x^4, x^3$ when $d$ is positive and the ordering $x^3, x^4, x^2$ when $d$ is negative.

Now solve each system of equations, starting with case 1:

$\begin{aligned} x^2 + d &= x^3 \\ d &= x^3 - x^2 \\ \implies x^2 + 2(x^3 - x^2) &= x^4 \\ -x^4 + 2x^3 - x^2 &= 0 \\ x^4 - 2x^3 + x^2 &= 0 \\ x^2(x-1)^2 &= 0 \\ x &= 0, 1 \end{aligned}$ However, neither of these solutions are valid, since every term in the sequence must be distinct.

Case 2: $\begin{aligned} x^2 + d &= x^3 \\ d &= x^3 - x^2 \\ \implies x^2 - (x^3 - x^2) &= x^4 \\ -x^4 - x^3 + 2x^2 &= 0 \\ x^4 + x^3 - 2x^2 &= 0 \\ x^2(x-1)(x+2) &= 0 \\ x &= 0, 1, -2 \end{aligned}$ The only valid solution here is $x=-2$

Case 3: $\begin{aligned} x^2 + d &= x^4 \\ d &= x^4 - x^2 \\ \implies x^2 + 2(x^4 - x^2) &= x^3 \\ 2x^4 - x^3 - x^2 &= 0 \\ x^2(x-1)(2x+1) &= 0 \\ x &= 0, 1, -\frac{1}{2} \end{aligned}$ Again, this gives us no valid solutions since none of these values of $x$ give a sequence of distinct integers.

Thus, our only possible sequences come from case 2, with $x=-2$ . This gives us two sequences, whose first three terms are:

$(-2)^4, (-2)^2, (-2)^3$ , with $d=-12$

$(-2)^3, (-2)^2, (-2)^4$ , with $d=12$

It can be seen that every number in the form $(-2)^2 + 12k$ for some integer $k$ must be a member of either of these sequences. So, if integer values of $k$ can be found such that $(-2)^2 + 12k = (-2)^5, (-2)^6, (-2)^7, \text{or } (-2)^8$ , then we know that they can appear later in either sequence. This can be checked fairly easily:

$\begin{aligned} (-2)^2 + 12k &= (-2)^5, (-2)^6, (-2)^7, (-2)^8 \\ 4 + 12k &= -32, 64, -128, 256 \\ 12k &= -36, 60, -132, 252 \\ k &= -3, 5, -11, 21 \end{aligned}$

All values of $k$ turn out to be integers, so $(-2)^5, (-2)^6, (-2)^7, \text{and } (-2)^8$ can ALL appear later in either sequence.

• If $x^3$ is the middle term, we have $2x^3=x^2+x^4$ . Dividing both sides by $x^2$ and rearranging we have $x^2-2x+1=(x-1)^2=0$

Therefore $x=1$ and all the terms in the A.P. are $1$ which is not possible as all terms are distinct.

• If $x^2$ is the middle term, we have $2x^2=x^3+x^4$ . Dividing both sides by $x^2$ and rearranging we have $x^2+x-2=0$ .

Therefore $x=1$ or $x=-2$ . $x=1$ is not possible(same reason as the previous case). If $x=-2$ , we must have $x^3$ as the first term as $x^3$ is negative and $x^2,x^4$ are positive. Therefore the terms of the A.P are $-8,4,16,..$

We can observe that $x^6$ and $x^8$ are terms of this sequence.

• If $x^4$ is the middle term, we have $2x^4=x^2+x^3$ . Dividing both sides by $x^2$ and rearranging we have $2x^2-x-1=0$ .

Therefore we have $x=1$ or $x=-0.5$ . $x=1$ is not possible(same reason as first case). If $x=-0.5$ , we must we must have $x^3$ as the first term as $x^3$ is negative and $x^2,x^4$ are positive. Therefore the terms of the A.P are $-0.125,0.0625,0.25,...$