• # question_answer 2) If the ${{p}^{th}}$ term of an A. P. is q and ${{q}^{th}}$ term is p, prove that its ${{n}^{th}}$ term is $(p+q-n)$.

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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