A) zero
B) \[\frac{p+q}{pq}\]
C) 1
D) \[\frac{pq}{p+q}\]
Correct Answer: C
Solution :
\[{{T}_{p}}\]of \[AP=\frac{1}{q}\Rightarrow A+(p-1)d=\frac{1}{q}\] ?(i) \[{{T}_{q}}\]of \[AP=\frac{1}{p}\Rightarrow A+(q-1)\,d=\frac{1}{p}\] ?(ii Subtracting \[d(p-1-q+1)=\frac{1}{q}-\frac{1}{p}\] \[d(p-q)=\frac{p-q}{pq}\Rightarrow d=\frac{1}{pq}\] from (i) \[A+(p-1)\frac{1}{pq}=\frac{1}{q}\] \[A+\frac{p}{pq}-\frac{1}{pq}=\frac{1}{q}\Rightarrow A=\frac{1}{pq}\] \[{{T}_{pq}}\]of \[AP=A+(pq-1)d\] \[\Rightarrow \] \[=\frac{1}{pq}+\frac{pq}{pq}-\frac{1}{pq}=1\] \[{{T}_{pq}}\]of \[H.P.=1\]
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