If p th term of an AP is q and q th term is p, then show that its n th

1.	If p th term of an AP is q and q th term is p, then show that its n th term is (p+q-n)
Answer» Let \'a\' be the first term and d be the common difference of the given A.P.Then,according to question we have\xa0Tp = a + ( p - 1 )d and Tq = a + ( q - 1 )d.(where Tp\xa0and Tq\xa0are pth\xa0and qth\xa0terms\xa0of given A.P)Now, Tp = q and Tq=\xa0p (given).a + (p -1)d = q ...(i)and a + (q - 1) d = p ...(ii)On subtracting (i) from (ii), we get{tex}\\implies{/tex}(q-1)d-(p-1)d=p-q{tex}\\implies{/tex}qd-d-pd+d =p-q{tex}\\implies{/tex}(q - p)d = ( p - q ){tex}\\implies d=\\frac{p-q}{q-p}{/tex}{tex}\\Rightarrow{/tex}d = -1.Putting d = -1 in (i), we get a = (p + q -1).Thus, a = (p + q - 1 ) and d = -1.Therefore, nth term = a+ (n- 1)d = (p + q -1) + (n -1) {tex}\\times{/tex}\xa0(-1)=p+q-1-n+1\xa0= p + q - n.Hence, nth term = (p + q - n). Which is required answer.

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