In the expansion of $(1+x)^n$, the general term i.e. $(r+1)$ th tem is
$$
T_{Y+1}={ }^n C_Y(1)^{n-y} X^Y={ }^n C_Y X^Y
$$
$\therefore$ Coefficient of $(r+1)$ th term is ${ }^n C_Y$
Similarly, coefficient of $p$ th term $={ }^n C_{p-1}$
$$
\begin{aligned}
\therefore \quad p & ={ }^n C_{p-1} \text { (given) } \\
p & =\frac{n !}{(p-1) !(n-P+1) !}
\end{aligned}
$$
and coefficient of $\left(p+1\right.$ th term $={ }^n C_p$
$\therefore \quad q={ }^n C_p$ (given)
On dividing Eq. (i) by Eq. (ii), we get
$$
\begin{aligned}
& \frac{p}{q}=\frac{\frac{n !}{(p-1) !(n-p+1) !}}{{ }^n C_p}=\frac{\frac{n !}{(p-1) !(n-p+1) !}}{\frac{n !}{(p) !(n-p) !}} \\
& =\frac{(p) !(n-p) !}{(n-p+1) !(p-1) !}=\frac{p(p-1) !(n-p) !}{(n-p+1)(n-p) !(p-1) !}
\end{aligned}
$$


$$
\begin{array}{lc}
\Rightarrow & \frac{p}{q}=\frac{p}{n-p+1} \\
\Rightarrow & \frac{1}{q}=\frac{1}{n-p+1} \\
\Rightarrow & n-p+1=q \\
\Rightarrow & p+q=n+1
\end{array}
$$