A) \[\frac{pq-1}{2}\]
B) \[\frac{1-pq}{2}\]
C) \[\frac{pq+1}{2}\]
D) \[-\frac{pq+1}{2}\]
Correct Answer: C
Solution :
[c] Since \[{{T}_{p}}=a+(p-1)d=\frac{1}{q}....(i)\] and \[{{T}_{q}}=a+(q-1)d=\frac{1}{p}....(ii)\] From (i) and (ii), we get \[a=\frac{1}{pq}\] and \[d=\frac{1}{pq}\] Now sum of pq terms \[=\frac{pq}{2}\left[ \frac{2}{pq}+(pq-1)\frac{1}{pq} \right]\] \[=\frac{pq}{2}.\frac{2}{pq}\left[ 1+\frac{1}{2}(pq-1) \right]=\left[ \frac{2+pq-1}{2} \right]=\frac{pq+1}{2}\]
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