3.55 A retired worker receives an annual pension, the amount of which is proportional to the square root of the number of years he has served. If he had served $a$ more years, his pension would be $p$ yuan more than his original pension, and if he had served $b$ more years $(b \neq a)$, his pension would be $q$ yuan more than his original pension. Then, his annual pension is (expressed in terms of $a, b, p, q$)
(A) $\frac{p^{2}-q^{2}}{2(a-b)}$.
(B) $\frac{(p-q)^{2}}{2 \sqrt{a b}}$.
(C) $\frac{a p^{2}-b q^{2}}{2(a p-b q)}$.
(D) $\frac{a q^{2}-b p^{2}}{2(b p-a q)}$.
(E) $\sqrt{(a-b)(p-q)}$.
(21st American High School Mathematics Examination, 1970)
［Solution］Let $x$ be the annual pension, $y$ be the years of service, and $k$ be the proportionality constant. According to the problem, we have
$$
\left\{\begin{array}{l}
x = k \sqrt{y}, \\
x + p = k \sqrt{y + a}, \\
x + q = k \sqrt{y + b};
\end{array} \text{ or } \quad \left\{\begin{array}{l}
x^{2} = k^{2} y, \\
(x + p)^{2} = k^{2} (y + a), \\
(x + q)^{2} = k^{2} (y + b).
\end{array}\right.\right.
$$

Substituting (1) into (2) and (3) and simplifying, we get $2 p x + p^{2} = k^{2} a$ and $2 q x + q^{2} = k^{2} b$.
Thus, $\frac{2 p x + p^{2}}{2 q x + q^{2}} = \frac{a}{b}$, solving for $x$ gives $x = \frac{a q^{2} - b p^{2}}{2 (b p - a q)}$, where $b p \neq a q$.
When $b p = a q$, multiplying the above two equations by $q$ and $p$ respectively and then subtracting them yields
$$
p q (p - q) = k^{2} (a q - b p) = 0,
$$

which means
$$
p q = 0 \text{ or } p - q = 0.
$$

If $p q = 0$, then $p = 0$ or $q = 0$, which implies that $a$ or $b$ is 0, making one of the equations (2) or (3) always equal to equation (1);
If $p - q = 0$, i.e., $p = q$, then $a = b$, which contradicts the assumption.
In summary, the correct choice is $(D)$.