We claim that if $n = 2x$, the answer is $x^2 - x$. If $n = 2x + 1$, the answer is $x^2$. We prove both statements by induction; assume that the first statement holds for $n = 2x$. Then, we add another term, $a_{2x + 1}$. The possible choices for the 1st term of the three-term A.Ps are $a_{2x - 1}$, $a_{2x - 3}$, ..., $a_1$, so $x$ new A.Ps are added. Then, we add another term, $a_{2x + 2}$. The possible choices for the 1st term of the three-term A.Ps are $a_{2x}$, $a_{2x - 2}$, ..., $a_2$, so $x$ new A.Ps are added. We should have $x^2 + x$ A.P.s for $n = 2x + 2$. Plugging in $x + 1$ for $x$ in $x^2 - x$ we get $x^2 + x$, so the inductive hypothesis holds for $n = 2x$. Assume that the second statement holds for $n = 2x + 1$. Then, we add another term, $a_{2x + 2}$. As previously demonstrated, $x$ new A.Ps are added. Then, we add another term, $a_{2x + 3}$. The possible choices for the 1st term of the three-term A.Ps are $a_{2x + 1}$, $a_{2x - 1}$, ..., $a_1$, so $x + 1$ new A.Ps are added. We should have $x^2 + 2x + 1$ A.Ps for $n = 2x + 3$. This equals $(x + 1)^2$, so the inductive hypothesis holds for $n = 2x + 1$.