A) \[mn+1\]
B) \[\frac{mn+1}{2}\]
C) \[\frac{mn-1}{2}\]
D) \[\frac{mn-1}{3}\]
Correct Answer: B
Solution :
We have given, \[{{a}_{m}}=\frac{1}{n}\] and \[{{a}_{n}}=\frac{1}{m}\] Then, \[a+(m-1)d=\frac{1}{n}\] .....(i) And, \[a+(n-1)d=\frac{1}{m}\] ......(ii) Subtracting (ii) from (i), we get, \[\Rightarrow \] \[\frac{1}{n}-\frac{1}{m}=(m-n)d\,\,\,\Rightarrow \,\,d=\frac{1}{mn}\] \[\therefore \] \[\frac{1}{n}=a+(m-1)\frac{1}{mn}\] [From (i)] \[\Rightarrow \] \[\frac{1}{n}-\frac{m-1}{mn}=a\] \[\Rightarrow \] \[a=\frac{1}{mn}\] Now, \[{{S}_{mn}}=\frac{mn}{2}[2a+(mn-1)d]\] \[=\frac{1}{2}[mn+1]=\frac{mn+1}{2}\]
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