In an A.P., if pth term is 1q and qth term is 1p, prove that the sum of first pq terms is 12(pq+1), where p≠q.
General term of an A.P. is given by
an=a+(n−1)d
Sum of n terms of an A.P. is given by
Sn=n22a+(n−1)d
where a is the first term and d is the common difference.
Given ap=1q
⇒a+(p−1)d=1q ...(1)
aq=1p
⇒a+(q−1)d=1p ...(2)
Subtracting eq(2) from eq (1), we get
(p−1)d−(q−1)d=1q−1p
⇒(p−1−q+1)d=p−qpq
⇒(p−q)d=p−qpq
⇒d=1pq
Putting the value of d in (1), we get
a+(p−1)1pq=1q
⇒a=1q−1q+1pq ⇒a=1pq
Sum of pq terms of an A.P. is
Spq=pq2[2a+(pq−1)d]
=pq2[2pq+(pq−1)1pq]
=1+12(pq−1)
=12pq+1−12
=12pq+12
=12(pq+1)
So, the sum of pq terms of A.P. is 12(pq+1)