question_answer

If \[{{m}^{th}}\] term of an A.P. is \[\frac{1}{n}\] and \[{{n}^{th}}\] term is \[\frac{1}{m}\], then find the sum of its first mn terms.

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

\[=\frac{mn}{2}[2a+(mn-1)d]\]

\[=\frac{mn}{2}\left[ \frac{2}{mn}+(mn-1)\frac{1}{mn} \right]\]   \[\left[ \because \,\,a=\frac{1}{mn},d=\frac{1}{mn} \right]\]

\[=\frac{mn}{2}\left[ \frac{1}{mn}+1 \right]\]

\[=\frac{1+mn}{2}\]