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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 $a=\frac {1}{mn}$

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A question about the same AP was asked very recently. It is different from yours in that you are asking "where is the mistake." –  André Nicolas Jul 24 '13 at 15:57
    
@AndréNicolas yes that was the one I was trying to solve –  Ovi Jul 24 '13 at 15:58
    
why should it be that $mx = nx$ implies $m = n$? It could be $x = 0$? –  W_D Jul 24 '13 at 16:00
    
@AlexWhite Yea your right I'm assuming that $x \not =0$ but then using this result I get $x=0$ –  Ovi Jul 24 '13 at 16:02
1  
What @AlexWhite says is the reason. In fact, given that $mx = nx$ and you know that $m \neq n$ (because otherwise there is simply not enough data to answer the question), we can conclude that $x = 0$, necessarily. In fact, you can't get $a = \frac1{mn}$ without assuming that $m \neq n$. So if $x = 0$, then you can't conclude from $mx = nx$ that $m = n$; that conclusion is not valid. –  ShreevatsaR Jul 24 '13 at 16:04
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1 Answer

up vote 1 down vote accepted

There is a mistake that disappears. From $nx=mx$ you conclude $n=m$, which you cannot unless you know $x \neq 0$. Immediately after, you conclude $x=0$, which is all you use afterward. As others have remarked in the comments, you have to assume $m \neq n$ 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.

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Ohh the question asked for the sum! I understood that $m \not=n$ but I was bothered because I thought the question asked about the $mn^{th}$ term which is equal to $1$ . –  Ovi Jul 24 '13 at 16:19
    
@AndréNicolas: I had missed that in the middle. I have updated the answer. Thanks –  Ross Millikan Jul 24 '13 at 17:08
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