question_answer

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