Terms of the sequence

Ann, Kristina, and Ulla solved the following assignment:

A sequence $(a_n )$ is for each $n\in \mathbb{N}$ given by the formula $a_n=\frac{5-3n}{7}$. Find all values of $x\in \mathbb{R}$ so that $$ a_4,~x^2+2, ~a_{11} $$ are three consecutive terms of a geometric sequence and calculate its common ratio $q$.

Ann first calculated: $$ a_4=-1,~a_{11}=-4 $$ As for her, if the numbers $a_4$, $x^2+2$, $a_{11}$ are to be consecutive terms of a geometric sequence, then $$ x^2+2=\frac{a_{11}}{a_4} $$ and so $$ x^2+2=4 $$ Then, she solved the above equation and got: $$ x=\pm \sqrt{2} $$ The three consecutive terms of the sequence are $-1$, $4$, $-4$ and the common ratio of the sequence $q=4$.

Kristina remembered that if we consider three consecutive terms in a geometric sequence $a$, $b$, $c$, then $$ \frac{b}{a}=\frac{c}{b} $$ and $$ b^2=ac $$ She calculated the outer terms: $$ a_4=-1,~a_{11}=-4 $$ and proceeded this way: $$ \begin{gather} (x^2+2)^2=4 \cr |x^2+2|=2 \cr x^2+2=\pm2 \end{gather} $$ Solving these two equations, she got three results: $$ x=0,x=2,x=-2 $$ For $x=0$, the three consecutive terms are $-1$, $2$, $-4$ and $q=-2$.

For $x=\pm 2$, the three consecutive terms are $-1$, $4$, $-4$ and $q=4$.

Ulla thought that for the middle term in a geometric sequence must hold: $$ x^2+2=\frac{a_4+a_{11}}{2} $$ She calculated: $$a_4=-1,~a_{11}=-4 $$ hence $$ \begin{gather} x^2+2=-\frac{5}{2} \cr x^2=-\frac{9}{2} \end{gather} $$ From there, she concluded that the assignment had no solution.

Which one of them proceeded correctly in solving?