Question

If mth term of an AP is $\frac{1}{n}$ and nth term is $\frac{1}{m}$ then find the sum of its first mn terms. [CBSE 2017]

Answer 1

Suppose a be the first term and d be the common difference of the given AP.

$$\begin{gathered} a_m=\frac{1}{n} \\ \Rightarrow a+\left(m-1\right)d=\frac{1}{n}.....\left(1\right) \end{gathered}$$
And,
$$\begin{gathered} a_n=\frac{1}{m} \\ \Rightarrow a+\left(n-1\right)d=\frac{1}{m}.....\left(2\right) \end{gathered}$$
Subtracting (2) from (1), we get
$$\begin{gathered} \frac{1}{n}-\frac{1}{m}=\left(m-n\right)d \\ \Rightarrow \frac{m-n}{mn}=\left(m-n\right)d \\ \Rightarrow d=\frac{1}{mn} \end{gathered}$$
Putting $d=\frac{1}{mn}$ in (1), we get
$$\begin{gathered} a+\left(m-1\right)\frac{1}{mn}=\frac{1}{n} \\ \Rightarrow a+\frac{1}{n}-\frac{1}{mn}=\frac{1}{n} \\ \Rightarrow a=\frac{1}{mn} \end{gathered}$$
∴ Sum of mn terms,
$$\begin{gathered} S_{mn}=\frac{mn}{2}\left[2a+\left(mn-1\right)d\right] \\ =\frac{mn}{2}\left[\frac{2}{mn}+\left(mn-1\right)\frac{1}{mn}\right] \\ =\frac{mn}{2}\left(\frac{1+mn}{mn}\right) \\ =\left(\frac{mn+1}{2}\right) \end{gathered}$$