A) \[\frac{r}{mn}\]
B) \[\frac{mn}{r+1}\]
C) \[\frac{mn}{r}\]
D) \[\frac{mn}{r-1}\]
Correct Answer: C
Solution :
Given \[{{T}_{m}}=n,\ {{T}_{n}}=m\] for H.P. Therefore for the corresponding A.P. \[{{m}^{th}}\] term \[=\frac{1}{n},\ {{n}^{th}}\] term \[=\frac{1}{m}\] Let \[a\] and d be the first term and common difference of this A.P., then \[a+(m-1)d=\frac{1}{n}\] ?..(i) \[a+(n-1)d=\frac{1}{m}\] ?..(ii) Solving these, we get \[a=\frac{1}{mn},\ d=\frac{1}{mn}\] Now, \[{{r}^{th}}\]term of corresponding A.P. \[=a+(r-1)d=\frac{1}{mn}+(r-1)\frac{1}{mn}=\frac{1+r-1}{mn}=\frac{r}{mn}\] Therefore \[{{r}^{th}}\] term of corresponding H.P. is \[\frac{mn}{r}\]. Note: Students should remember this question as a fact.
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