The $m^{th}$ term of an A.P is ${}^{\prime }{n}^{\prime }$ and $n^{th}$ term is ${}^{\prime }{m}^{\prime }.$ Then $(m+n{)}^{th}$ term of it is

The correct option is A: $0$

We know that the $n^{th}$ of an AP is given by the formula $a_n=a+(n-1)d$

$m^{th}$ term $=a_m=a+(m-1)d=n\dots \dots \left(i\right)$ [$\because a_m=n$]
$n^{th}$ term $=a_n=a+(n-1)d=m\dots \dots \left(ii\right)$ [$\because a_n=m$]
Subtracting $eq.\left(ii\right)$ from $eq.\left(i\right)$, we get
$n-(m-1)d=m-(n-1)d$

$\Rightarrow (n-m)-md+d=-nd+d$

$\Rightarrow (n-m)=-nd+md$

$\Rightarrow d=\frac{n-m}{-(n-m)}$
$d=-1$

On putting the value of ${}^{\prime }{d}^{\prime }$ in $eq.\left(i\right),$ we have

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

$\Rightarrow a-m+1=n$

$\Rightarrow a=n+m-1$

So, $(m+1{)}^{th}$ term, $a_{m+n}=a+(m+n-1)d$

$=(n+m-1)+(m+n-1)(-1)=0$

$=n+m-1-m-n+1=0$