5. In the arithmetic sequence $\left\{a_{n}\right\}$, if $a_{3}+a_{4}+a_{10}+a_{11}=2002$, then $a_{1}+a_{5}+a_{7}+a_{9}+$ $a_{13}=$ $\qquad$
5. $2502 \frac{1}{2}$.
$$
a_{3}+a_{11}=2 a_{7}=a_{4}+a_{10}, a_{3}+a_{4}+a_{10}+a_{11}=4 a_{7}=2002, a_{7}=\frac{1}{2} \cdot 1001=500 \frac{1}{2} \text{. }
$$

And $a_{1}+a_{13}=a_{5}+a_{9}=2 a_{7}, a_{1}+a_{5}+a_{7}+a_{9}+a_{13}=5 a_{7}=2502 \frac{1}{2}$.