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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, sorry.. now it's okay – Gamma Oct 29 '12 at 16:26
    
What you said is $$ma_m=na_m\Longrightarrow n=m$$ unless $\,a_m=0\,$... Are you sure this is what you meant? – DonAntonio Oct 29 '12 at 16:27
    
So, what do you think what it should be... – Gamma Oct 29 '12 at 16:30
    
Ok, I think it should be my writing mistake. – Gamma Oct 29 '12 at 16:32
up vote 3 down vote accepted

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 $a=d$ – Gamma Oct 30 '12 at 0:47

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