7. If we denote $A(p, q, n)=|\dot{p}+j q| i, j \geqslant 0, i+j \leqslant n$, then the number of elements in $A(p, q, n)$ is
$$|A(p, q, n)|=\left\{\begin{array}{ll} \frac{(n+1)(n+2)}{2}, & n < p; \\ \frac{p(2 n-p+3)}{2}, & n \geqslant p. \end{array}\right.$$

According to the definition, we have
$$\begin{array}{l} A(p, q, n) \backslash A(p, q, n-1) \\ =\{i p+(n-i) q \mid i=1,2, \cdots, n\}. \end{array}$$

Notice that $i p+(n-i) q=p(i+q)+q(n-p-i)$, and
$(i+q)+(n-p-i)=n-p+q \leqslant n-1$, then
$$i p+(n-i) q \in A(p, q, n-1) \Leftrightarrow n-p-i \geqslant 0.$$

Therefore, $A(p, q, n) \backslash A(p, q, n-1)$
$$=\left\{\begin{array}{l} |\dot{p}+(n-i) q| i=n-p+1, n-p+2, \cdots, n \mid, n \geqslant p; \\ |\dot{p}+(n-i) q| i=0,1, \cdots, n \mid, n < p. \end{array}\right.$$

Let $a_{n}=|A(p, q, n)|$, then
$$a_{n}-a_{n-1}=\left\{\begin{array}{cc} p, & n \geqslant p; \\ n+1, & n < p. \end{array}\right.$$

Noting that $a_{0}=1$, for $n < p$, we have
$$\begin{array}{l} a_{n}=a_{0}+\left(a_{1}-a_{0}\right)+\cdots+\left(a_{n}-a_{n-1}\right) \\ =1+2+\cdots+(n+1)=\frac{(n+1)(n+2)}{2}; \end{array}$$

Therefore, for $n \geqslant p$, we have
$$\begin{array}{l} a_{n}=a_{p-1}+\left(a_{p}-a_{p-1}\right)+\cdots+\left(a_{n}-a_{n-1}\right) \\ =a_{p-1}+(n-p+1) p=\frac{p(2 n-p+3)}{2}. \end{array}$$