3. Let $A B$ be the major axis of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. If $A B$ is divided into 100 equal parts, and perpendicular lines are drawn through each division point to $A B$, intersecting the upper half of the ellipse at $P_{1}, P_{2}, \cdots \cdots, P_{99}$. $F_{1}$ is the left focus of the ellipse, then the value of $\left|F_{1} A\right|+\left|F_{1} P_{1}\right|+\left|F_{2} P_{2} i+\cdots+\right| F_{1} P_{99}|+| F_{1} B \mid$ is ( ).
A. $98 a$
B. $99 a$
C. $100 a$
D. $101 a$
3. D

From the definition of an ellipse, $\left|F_{1} P_{\mathrm{f}}\right|+\left|F_{2} P_{\mathrm{i}}\right|=2 a(i=1,2 \cdots, 99)$
Therefore, $\sum_{i=1}^{\infty}\left(1 F_{1} P_{i}|+| F_{2} P, \mid\right)=99 \times 2 a=198 a$, From the problem, we know that $P_{1}, P_{2}, \cdots \cdots$, $P_{99}$, are symmetrically distributed about the $y$-axis, so $\sum_{i=1}^{\infty}\left(\left|F_{1} P_{i}\right|\right)=\frac{1}{2} \sum_{i=1}^{\infty}\left(\left|F_{1} P_{i}\right|+\right.$ $\left.\left|F_{2} P_{i}\right|\right)=99 a$, and since $\left|F_{1} A\right|+\left|F_{i} B\right|=2 a$, the required value is $101 a$, hence the answer is $\mathrm{D}$.