The perpendicular distance from origin $(0, 0)$ to $\frac{x}{a} + \frac{y}{b} - 1 = 0$ is $p = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.
Squaring and inverting: $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} = \frac{a^2 + b^2}{a^2 b^2}$.
Since $a^2, p^2, b^2$ are in A.P., $2p^2 = a^2 + b^2 \implies p^2 = \frac{a^2 + b^2}{2}$.
Substituting $p^2$: $\frac{2}{a^2 + b^2} = \frac{a^2 + b^2}{a^2 b^2} \implies (a^2 + b^2)^2 = 2a^2 b^2 \implies a^2 + b^2 = \sqrt{2}ab$.
Thus, $a^2 + b^2$ is not equal to $ab$, $2ab$, or $a^2 b^2$.
(d)