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If $n$ times the $m^{th}$ term of an arithmetic progression is equal to $m$ times the $n^{th}$ term, find the $(m + n)^{th}$ term.

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    Ok, I think it should be my writing mistake.2012-10-29

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In an arithmetic progression $a,a+d,a+2d,...$, the $k$th term is $a+(k-1)d$. Hence $n(a+(m-1)d)=m(a+(n-1)d)$ so that $(n-m)(a-d)=0$. If we assume $n\neq m$, then we have $a=d$. This means that the arithmetic progression is $a,2a,3a,...$ so that the $(m+n)$th term is $(m+n)a$.

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    How assuming $n\neq m$ makes $n(a-d)=m(a-d)$ to $a=d$2012-10-30