A sequence with variables, find the $mn^{th}$ term given the $m^{th}$ and the $n^{th}$? I was trying to answer this question, but I'm getting a wrong answer. The question is: If the $m$th term of an arithmetic progression is $\frac{1}{n}$ and the $n$th term is $\frac{1}{m}$ then prove that the sum to $mn$ terms is $\frac{mn+1}{2}$ Let's say the sequence goes like this: $x+a, x+2a, x+3a, ...$ so that the $n^{th}$ term is $x+na=\frac 1m$ and the $m^{th}$ term is $x+ma=\frac 1n$ . We divide the equations such that $\frac {x+ma}{x+na}=\frac mn$ $\to$ $nx+mna=mx+mna$ , so $m=n$ . Therefore, $x+\frac 1m =\frac 1m$ , so $x=0$ Then, $0+ma=\frac 1n$ , so $a=\frac {1}{mn}$ The $(mn)^{th}$ term is $x+mna$, which is equal to $1$ . What is the mistake here? Thanks. EDIT There is another problem because is $m=n$ , division by $0$ works: If we subtract the two equations, we get $a(m-n)=\frac {m-n}{mn}$ and if we divide both sides by $0$ we get the correct answer of $a=\frac {1}{mn}$

There is a mistake that disappears. From $nx=mx$ you conclude $n=m$, which you cannot unless you know $x \
eq 0$. Immediately after, you conclude $x=0$, which is all you use afterward. As others have remarked in the comments, you have to assume $m \
eq n$ or you don't have enough information, then you can conclude $x=0$. Now given your result, the $i^{\text{th}}$ term is $\frac i{mn}$ and the sum of the first $mn$ terms is $\frac {mn(mn+1)}{2mn}=\frac {mn+1}2$ as desired.