Problem-3

Question

Let \(m\) and \(n\) be two distinct natural numbers. If the \(\displaystyle m^{th}\) term of an AP is \(\displaystyle \cfrac{1}{n}\) and the \(\displaystyle n^{th}\) term is \(\displaystyle \cfrac{1}{m}\), show that the sum of the first \(\displaystyle mn\) terms of the AP is \(\displaystyle \cfrac{1}{2}( mn+1)\).

Solution

Let the first term be \(\displaystyle a\) and the common difference be \(\displaystyle d\).

\[ \begin{equation*} \begin{aligned} \cfrac{1}{m} & =a+( n-1) d\\ & \\ \cfrac{1}{n} & =a+( m-1) d \end{aligned} \end{equation*} \tag{1}\]

Subtracting:

\[ \begin{equation*} \begin{array}{ c c l } & \cfrac{1}{m} -\cfrac{1}{n} & =( n-m) d\\ & & \\ \Longrightarrow & \cfrac{n-m}{mn} & =( n-m) d\\ & & \\ \Longrightarrow & d & =\cfrac{1}{mn} \end{array} \end{equation*} \]

Plugging \(\displaystyle d=\cfrac{1}{mn}\) back into Equation 1:

\[ \begin{equation*} \begin{array}{ c c l } & \cfrac{1}{m} & =a+( n-1) d\\ & & \\ \Longrightarrow & a & =\cfrac{1}{m} -\cfrac{( n-1)}{mn}\\ & & \\ \Longrightarrow & a & =\cfrac{1}{mn} \end{array} \end{equation*} \]

Therefore, we see that \(\displaystyle a=d=\cfrac{1}{mn}\). With this, the sum of the first \(\displaystyle mn\) terms is:

\[ \begin{equation*} \begin{array}{ c c l } & S_{mn} & =\cfrac{mn}{2}[ 2a+( mn-1) d]\\ & & \\ & & =\cfrac{mn}{2}\left[\cfrac{2}{mn} +\cfrac{mn-1}{mn}\right]\\ & & \\ & & =\cfrac{1}{2}( mn+1) \end{array} \end{equation*} \]