Geometric and arithmetic sequence II

Piotr and Paweł were to solve the following task:

A sequence $(a_n )$ is defined recursively by: $$ \begin{align} a_1&=2 \cr a_n&=3(n-1)-a_{n-1}+3,~n\geq 2 \end{align} $$

Find integers $x$, $y$ such that $a_7$, $a_2+x$, $a_4+2y$ are three consecutive terms of an arithmetic sequence, and $a_1$, $4x-a_2$, $3a_6-y$ are three consecutive terms of a geometric sequence.

They both calculated the following six terms of the sequence: $$ \begin{align} a_2&=3-a_1+3=4 \cr a_3&=6-a_2+3=5\cr a_4&=9-a_3+3=7\cr a_5&=12-a_4+3=8\cr a_6&=15-a_5+3=10\cr a_7&=18-a_6+3=11 \end{align} $$

Piotr continued this way: If $a_7$, $a_2+x$, $a_4+2y$ are three consecutive terms of an arithmetic sequence, then the difference between any two consecutive terms is always the same, and if $a_1$, $4x-a_2$, $3a_6-y$ are three consecutive terms of a geometric sequence, then the middle term can be calculated in terms of the outer two terms: $$ \begin{align} (4+x)-11&=(7+2y)-(4+x)\cr (4x-4)^2&=2(30-y)
\end{align} $$ He expressed $x$ from the first equation: $$ \begin{align} 2x&=10+2y \cr x&=5+y \end{align} $$ and substituted it into the second equation: $$ \begin{align} (4y+16)^2&=60-2y \cr 16y^2+130y+196&=0 \cr 8y^2+65y+98&=0 \end{align} $$ Finally, he solved the quadratic equation that he got: $$ \begin{align} y_{1,2}&=\frac{-65 \pm \sqrt{65^2-4 \cdot 8 \cdot 98}}{16} \cr y_{1,2}&=\frac{-65 \pm \sqrt{4225-3136}}{16}\cr y_{1,2}&=\frac{-65 \pm \sqrt{1089}}{16}\cr y_{1,2}&=\frac{-65 \pm 33}{16}\cr y_1&=-2,y_2=-\frac{49}{8} \end{align} $$ Now, all that was left to do was to calculate $x$: $$ x_1=3,~x_2=-\frac98 $$ He thus obtained two solutions: $x_1=3$, $y_1=-2$ and $x_2=-\frac98$, $y_2=-\frac{49}{8}$.

Paweł reasoned this way: If $a_7$, $a_2+x$, $a_4+2y$ are three consecutive terms of an arithmetic sequence, then the middle term is an arithmetic mean of the outer two terms, and if $a_1$, $4x-a_2$, $3a_6-y$ are three consecutive terms of a geometric sequence, then the common ratio is the same for any two consecutive terms: $$ \begin{align} 4+x&=\frac12 (11+(7+2y)) \cr \frac{4x-4}2&=\frac{30-y}{4x-4} \end{align} $$

He expressed $y$ from the second equation: $$ \begin{align} (4x-4)^2&=60-2y \cr 16x^2-32x+16&=60-2y \cr y&=-8x^2+16x+22 \end{align} $$ and substituted it into the first equation and got: $$ \begin{align} 8+2x&=11+7-16x^2+32x+44 \cr 16x^2-30x-54&=0 \cr 8x^2-15x-27&=0 \end{align} $$ Then he solved the above equation: $$ \begin{align} x_{1,2}&=\frac{-15 \pm \sqrt{225-4 \cdot 8 \cdot (-27) }}{16} \cr x_{1,2}&=\frac{-15 \pm \sqrt{225+864}}{16}\cr x_{1,2}&=\frac{-15 \pm \sqrt{1089}}{16}\cr x_{1,2}&=\frac{-15 \pm 33}{16}\cr x_1&=-3,~x_2= \frac98 \end{align} $$

Finally, he determined the remaining unknown $y$: $$ \begin{align} y_1&=-8 \cdot 9-48+22=-98 \cr y_2&=-8 \cdot \frac{81}{64}+16 \cdot \frac{9}{8}+22=\frac{239}{8} \end{align} $$

As for him the task has one solution, and that is $x=-3$ and $y=-98$.

Which one of them proceeded correctly in solving?