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 $a_k=a_0+kp$. Then:
$a_n=a_m+(n-m)p\\
\frac{1}{m}-\frac{1}{n}=(n-m)p\\
\frac{n-m}{nm}\frac{1}{n-m}=p\\
p=\frac{1}{nm}$
and 
$a_n=a_0+np\\
a_0=a_n-np=\frac{1}{m}-\frac{1}{m}=0$
Now we just need to calculate
$\sum\limits_{k=0}^{nm}a_k=\sum\limits_{k=0}^{nm}a_0+kp= \frac{1}{nm}\sum\limits_{k=0}^{nm}k= \frac{1}{nm}\frac{(nm+1)(0+nm)}{2} = \frac{1}{2}(nm+1)$
