A) \[\frac{pq}{p+q}\]
B) \[\frac{{{p}^{2}}-{{q}^{2}}}{p+q}\]
C) \[\frac{{{p}^{2}}-{{q}^{2}}}{p-q}\]
D) \[\frac{{{p}^{2}}-{{q}^{2}}}{{{p}^{3}}-{{q}^{3}}}\]
Correct Answer: A
Solution :
(a): Let ?a? and ?d? be the first term and common difference of corresponding AP. i.e. we have \[\frac{1}{a},\frac{1}{a+d},......\frac{1}{a+[p-1]d}\]???., \[\frac{1}{a+(q-1)d}\] ??., \[\frac{1}{a+(p+q-1)d}\] as the sequence. The pth and qth terms of the AP are \[q=a+(p-1)d\] and \[p=a+(q-1)d\] For series to be in HP, we have \[\frac{1}{q}=a+(p-1)d\] and \[\frac{1}{p}=a+(q-1)d\] ?.(1) \[\Rightarrow \]\[\frac{1}{q}-\frac{1}{p}=(p-q)d\] \[\Rightarrow \] \[\frac{1}{pq}=d\] Putting this is (1), we will get a \[=\frac{1}{pq}\] Therefore, \[{{(p+q)}^{th}}\] term of HP \[\text{=}\frac{\text{1}}{\text{(p+q)}\,\,\text{th}\,\,\text{of}\,\,\text{AP}}\text{=}\frac{\text{1}}{\text{a+(p+q-1)d}}\]\[\text{=}\frac{1}{\frac{1}{pq}+\frac{(p+q-1)}{pq}}=\frac{pq}{p+q}\] Hence, \[{{(p+q)}^{th}}\] of the HP is given by \[\frac{pq}{p+q}\]
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