Answer:
Let a and d be the first term and common difference respectively of the given A.P. Then, \[\frac{1}{n}={{m}^{th}}\] term \[\Rightarrow \frac{1}{n}=a+(m-1)d\] ?(i) \[\frac{1}{m}={{n}^{th}}\] term \[\Rightarrow \frac{1}{m}=a+(n-1)d\] ?(ii) By subtracting eq. (ii) from eq. (i), \[\frac{1}{n}-\frac{1}{m}=(m-1)d\] \[\Rightarrow \] \[\frac{m-n}{mn}=(m-n)d\] \[\Rightarrow \] \[d=\frac{1}{mn}\] Putting \[d=\frac{1}{mn}\] in eq. (i) We get, \[\frac{1}{n}=a+(m-1)\frac{1}{mn}\] \[\Rightarrow \] \[\frac{1}{n}=a+\frac{1}{n}-\frac{1}{mn}\] \[\Rightarrow \] \[a=\frac{1}{mn}\] Sum of first mn terms \[=\frac{mn}{2}[2a+(mn-1)d]\] \[=\frac{mn}{2}\left[ \frac{2}{mn}+(mn-1)\frac{1}{mn} \right]\] \[\left[ \because \,\,a=\frac{1}{mn},d=\frac{1}{mn} \right]\] \[=\frac{mn}{2}\left[ \frac{1}{mn}+1 \right]\] \[=\frac{1+mn}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec