5. A group of two-digit numbers forms an even-numbered arithmetic sequence, the sum of all odd-numbered terms is 100. If starting from the 1st term, each odd-numbered term is merged with the adjacent even-numbered term that follows it without changing the order to form a new four-digit number, thus creating a new sequence, then the difference between the sum of the new sequence and the sum of the original sequence is $\qquad$
【Solution】Solution: Let the odd terms of this arithmetic sequence be $a_{1}, a_{3}, a_{5}, \cdots$, with a common difference of $d$. Then, by merging each odd term with the adjacent even term that follows, since each term is a two-digit number, the resulting four-digit sequence can be represented as $a_{1} \times 100+a_{1}+d, a_{2} \times 100+a_{2}+d, \cdots$,

Therefore, the sum of the new sequence differs from the sum of the original sequence by $99 \times\left(a_{1}+a_{3}+a_{5}+\cdots\right)$. Since the sum of the odd terms is 100, $99 \times\left(a_{1}+a_{3}+a_{5}+\cdots\right)=99 \times 100=9900$, so the answer is 9900.