If $m$th term and $n$th term of arithmetic sequence are $1/n$ and $1/m$ then the sum of the first $mn$ terms of the sequence is $(mn+1)/2$ 
If $m$th term and $n$th term of arithmetic sequence are $1/n$ and $1/m$ respectively then prove that the sum of the first $mn$ terms of the sequence is $(mn+1)/2$.

My Attempt ;
$$\textrm t_{m}=\dfrac {\textrm 1}{\textrm n}$$
$$\textrm a + \textrm (m-1)d =\dfrac {1}{n}$$
And, 
$$\textrm t_{n}=\dfrac {1}{m}$$
$$\textrm a+\textrm (n-1)d=\dfrac {1}{m}$$
What do I do further? 

I am assuming equations created by you as equation (1) and (2) respectively.
From equation (1),
$a = \frac 1n - (m - 1)d$
Put value of a in equation (2),
$\frac 1n - (m - 1)d + (n - 1)d = \frac 1m$
$\implies (-m + 1 + n - 1)d = \frac 1m - \frac 1n$
$$\implies (n - m)d = \frac {n - m}{mn}$$
$$\implies d = \frac 1{mn}$$
Then $$a = \frac 1n - (m - 1) \cdot \frac{1}{mn}$$
$$a = \frac 1n - \frac 1n + \frac 1{mn}$$
$$a = \frac 1{mn}$$
Using sum formula -
$S_{mn}=\frac{mn}2 \left[ 2 \cdot \frac {1}{mn} + (mn - 1) \cdot  \frac {1}{mn} \right]$
Taking $\frac {1}{mn}$ common,
$= \frac{mn}2 × \frac {1}{mn}\left[ 2 + (mn - 1) \right]$
$= \frac{1}2 × \left( mn + 1) \right)$
