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? 

Let $u_t = u_0 + a \cdot t$
$$ u_m = \frac{1}{n} $$
$$ u_n = \frac{1}{m} $$
Therefore $$ u_n - u_m = a \cdot \left( n-m \right)= \frac{1}{m} - \frac{1}{n} $$
$$ a=\frac{1}{m \cdot n} $$
So the initial term is$$ u_n = u_0 + \frac{n}{m \cdot n} = u_0 + \frac{1}{m} = \frac {1}{m}$$
$$ u_0 = 0 $$
We have $$ u_t = \frac{t}{m\cdot n} $$
We want to determine $$\sum_{t=0}^{m\cdot n} u_t = \sum_{t=0}^{m\cdot n} \frac{t}{m \cdot n}$$
$$\sum_{t=0}^{m\cdot n} u_t = \frac{1}{m \cdot n} \sum_{t=0}^{m\cdot n} t$$
$$\sum_{t=0}^{m\cdot n} u_t = \frac{1}{m \cdot n} \frac{(m \cdot n)\cdot(m \cdot n +1)}{2}$$
$$\sum_{t=0}^{m\cdot n} u_t = \frac{m \cdot n +1}{2}$$
