For $n\in N$ if two consecutive terms in the expansion of ${\left(p+q\right)}^n$  are equal then $\frac{\left(n+1\right)q}{p+q}$  is

${}^n{C}_r{p}^{n-r}{q}^r={}^n{C}_{r+1}{p}^{n-r-1}{q}^{r+1}$

$\frac{p^{n-r}{q}^r}{p^{n-r-1}{q}^{r+1}}=\frac{{}^n{C}_{r+1}}{{}^n{C}_r}$

$p^{n-r-n+r+1}{q}^{r-r-1}=\frac{n-r+1}{r}$

$p{q}^{-1}=\frac{n-r+1}{r}$

$\frac{p}{q}=\frac{n-r+1}{r}$

$\frac{p}{q}=\frac{n+1}{r}-\frac{r}{r}$

$\frac{\left(n+1\right)q}{p+q}=r$

$\frac{\left(n+1\right)}{r}=\frac{p}{q}+1=\frac{p+q}{q}$

$\therefore r$  is positive integer.