Two APs have the same common difference. The first term of one $AP$ is $2$ and that of the other is $7$. The difference between their $10^{\text{th}}$ terms is the same as the difference between their $21^{\text{st}}$ terms,which is the same as the difference between any two corresponding terms. Why?

(N/A) Let the common difference of both $APs$ be $d$.
Given that the first term of the first $AP$ is $a_1 = 2$ and the first term of the second $AP$ is $b_1 = 7$.
The $n^{\text{th}}$ term of an $AP$ is given by $a_n = a + (n-1)d$.
For the first $AP$,the $n^{\text{th}}$ term is $a_n = 2 + (n-1)d$.
For the second $AP$,the $n^{\text{th}}$ term is $b_n = 7 + (n-1)d$.
The difference between the $n^{\text{th}}$ terms is $b_n - a_n = [7 + (n-1)d] - [2 + (n-1)d] = 7 - 2 = 5$.
Since the difference is independent of $n$,the difference between any two corresponding terms is always $5$.
Thus,the difference between the $10^{\text{th}}$ terms is $5$,and the difference between the $21^{\text{st}}$ terms is also $5$.