Answer:
Let a be first term and d be common difference. Then, \[{{p}^{th}}\text{term}=q\Rightarrow a+(p-1)d=q\] ?(i) \[{{q}^{th}}\text{term}=p\Rightarrow a+(q-1)d=p\] ?(ii) On subtracting eq. (ii) from eq. (i) \[(p-1)d-(q-1)d=q-p\] \[pd-d-qd+d=q-p\] \[(p-q)d=q-p\] or \[d=\frac{q-p}{p-q}=-1\] Putting value of d in eq. (i) \[a+(p-1)(-1)=q\] \[a=q+p-1\] \[{{n}^{th}}term=a+(n-1)d\] \[=q+p-1+(n-1)(-1)\] \[=q+p-1+1-n=q+p-n\] \[{{T}_{n}}=q+p-n\] Hence Proved
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