Answer :

Given:


Now, ap = a + (p – 1)d



aq + q(p – 1)d = 1


aq + pqd – qd = 1 …(i)




ap + pqd – pd = 1 …(ii)


From eq. (i) and (ii), we get


aq + pqd – qd = ap + pqd – pd


aq – ap – qd + pd = 0


a(q – p) – d (q – p) = 0


(q – p)[a – d] = 0


a = d


Now, putting the value of a in eq. (i), we get


dq + pqd – qd = 1


pqd = 1



Hence,


Now, we have to show that sum of first pq term is


We know that,



For sum of first pq terms,


Putting n = pq,





Thus, sum of first pq terms is


Hence Proved


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