If $m^{\text{th}}$ term of an A.P. is $n$ and $n^{\text{th}}$ term is $m$, then its $10^{\text{th}}$ term is …
🎥 Video solution / Text Solution of this question is given below:
Let first term $a$, common difference $d$.
$a+(m-1)d=n \quad\text{and}\quad a+(n-1)d=m$.
Subtract: $(m-n)d=n-m \Rightarrow d=-1$.
Then $a=n+(m-1)=m+n-1$.
$t_{10}=a+9d=(m+n-1)+9(-1)=m+n-10$.