A) 0
B) 1
C) pq
D) \[pq(p+q)\]
Correct Answer: B
Solution :
Let a be the first term and d be the common difference of the corresponding A.P. \[{{p}^{th}}\]term of A.P. \[({{T}_{p}})=a+(p-1)d=\frac{1}{q}\] ?..(i) \[{{q}^{th}}\] term of A.P. \[({{T}_{q}})=a+(q-1)d=\frac{1}{p}\] ?..(ii) From (i) - (ii), \[(p-q)d=\frac{1}{q}-\frac{1}{p}=\frac{p-q}{pq}\,\,\Rightarrow \,d=\frac{1}{pq}\] From (i), \[a+(p-1)\frac{1}{pq}=\frac{1}{q}\] Þ \[a=\frac{1}{pq}\]. \\[{{T}_{pq}}=a+(pq-1)d=\frac{1}{pq}+(pq-1)\frac{1}{pq}=1\]. Therefore \[p{{q}^{th}}\] term is 1.
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