equals [IIT 1998]
A) \[\frac{1}{mn}\]
B) \[\frac{1}{m}+\frac{1}{n}\]
C) 1
D) 0
Correct Answer: C
Solution :
\[{{T}_{m}}=a+(m-1)\,d=\frac{1}{n}\] and \[{{T}_{n}}=a+(n-1)\,d=\frac{1}{m}\] On solving \[a=\frac{1}{mn}\] and \[d=\frac{1}{mn}\] \[\therefore \]\[{{T}_{mn}}=a+(mn-1)\,d=\frac{1}{mn}+(mn-1)\frac{1}{mn}=1\]
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