If mth term of an A.P. is n and nth term is m,

Given:

$a_{m}=n$

$\Rightarrow a+(m-1) d=n \quad \ldots(1)$

$a_{n}=m$

$\Rightarrow a+(n-1) d=m \quad \ldots .(2)$

Solving equations $(1)$ and $(2)$, we get:

$d=-1$

$a=n+m-1$

pth term:

$a_{p}=a+(p-1) d$

$=n+m-1+(p-1)(-1)$

$=n+m-p$

Hence, the pth term is $n+m-p$.