Let $a_1$ and $d$ be the first term and the difference of the arithmetic progression, respectively. From the condition $a_1$, $a_1 + 6d$ and $a_1 + 16d$ are consecutive members of a geometric progression, i.e.
$$(a_1 + 6d)^2 = a_1 \cdot (a_1 + 16d) \iff d \cdot (a_1 - 9d) = 0.$$
Since $d \neq 0$, we get that $a_1 = 9d$. Furthermore, we have $(x-9)(x+\sqrt{12-x}) = 0$ and $x \le 12$. Then $x = 9$ or $\sqrt{12-x} = -x$, i.e. $x^2 + x - 12 = 0$ and $x \le 0$, whence $x = -4$. Then $a_1 = 9$ and $a_1 = -4$, as $d = 1$ and $d = -\frac{4}{9}$, respectively. $\square$

Let $a_1$ and $d$ be the first term and the difference of the arithmetic progression, respectively. From the condition $a_1$, $a_1 + 6d$ and $a_1 + 16d$ are consecutive members of a geometric progression, i.e.
$$(a_1 + 6d)^2 = a_1 \cdot (a_1 + 16d) \iff d \cdot (a_1 - 9d) = 0.$$
Since $d \ne 0$, we get that $a_1 = 9d$. Furthermore, we have $(x-9)(x+\sqrt{12-x}) = 0$ and $x \le 12$. Then $x = 9$ or $\sqrt{12-x} = -x$, i.e. $x^2 + x - 12 = 0$ and $x \le 0$, whence $x = -4$. Then $a_1 = 9$ and $a_1 = -4$, as $d = 1$ and $d = -\frac{4}{9}$, respectively. $\square$