This question appeared on my test today, and two of the answer choices were
(a) $d=-1$ (b) $m=n$
and I am confused by some of the conclusions. We have $$T_m=n \implies a+(m-1)d=n \tag{1}$$ $$T_n=m \implies a+(n-1)d=m \tag{2}$$ Thus, subtracting the two equations from one another $$(m-n)d=n-m \implies d=-1$$ if $m \ne n$. But $$(1) \land (2) \implies a=n-(m-1)d=m-(n-1)d $$ $$\iff n-md+d=m-nd+d \iff m(d+1)=n(d+1) \implies m=n$$ if $d \ne -1$
Now, which one of the choices is more general or more accurate? I really thought hard but couldn't figure it out. Any help would be appreciated.
Your two equations $(m-n)d=n-m$ and $m(d+1)=n(d+1)$ are both factorized to $(m-n)(d+1)=0$.
My thought is that we cannot choose one between $m=n$ and $d=-1$.
For example, let $m=n$ and $(T_k)$ be a arithmetic sequence given by $T_k = m$. (i.e. $d=0, a=m$). Then this sequence satisfies the condition $T_m = n$, $T_n =m$. (You can produce other examples with $d \neq 0$) However, $d$ could differ from $-1$.
On the other hand, define $(T_k)$ by $T_k = m+n - k$ (i.e. $a=m+n-1, d=-1$) with given $m,n$ (they are not necessarily same). Then $T_m = n$ and $T_n$, but there is no guarantee that $m=n$.