Arithmetic Progression, Tn and Tm proving sum

Question

How can I prove this?

If in an A.P., Tm=n, Tn=m, prove d=-1.

Answer 1

Assume it is given that $m\ne n$.

You have $T(m)=n$ and $T(n)=m$. So,

$$a+(m-1)d=n\quad \text{and} \quad a+(n-1)d=m$$

Hint: Subtract the two equalities.

Answer 2

Notwithstanding the imprecision of the question, and assuming $m\neq n$:

The slope is $\frac{T_m-m}{T_n-n}=\frac{n-m}{m-n}=-1$

Answer 3

we have

$(T_i) $ arithmetic sequence

$$\implies T_n=T_m+(n-m)d $$

$$\implies m-n=(n-m)d $$

$$\implies (d+1)(n-m )=0$$

$$\implies d=-1$$