If the $m$th term of an Arithmetic Progression is $\frac{1}{n}$ and the $n$th term is............ Problem : 
If the $m$th term of an A.P is $\frac{1}{n}$ and the $n$th term is $\frac{1}{m}$ then prove that the sum to $mn$ terms is $\frac{mn+1}{2}$
My working : 
Let $a$ be the first term of the progression and $d$ the common difference then: 
$$\tag1T_m = \frac{1}{n}= a+(m-1)d$$ 
$$\tag2 T_n = \frac{1}{m} = a+(n-1)d$$
Subtracting (1) from (2) and solving for $d$ we get : 
$d = \frac{1}{mn}$ Please suggest what to do further.  Thanks 

Now that you have $d$ you can calculate $a$ from $(1)$
$$T_1 = a = \frac{1}{n}-(m-1)d = \frac{m}{mn}-\frac{m-1}{mn}=\frac{1}{mn}$$
Having $a$ and $d$ Apply the formula for arithmetic progression sum. 
$$T_{mn} = a + (mn-1)d = \frac{1}{mn} + \frac{mn-1}{mn} = 1$$
$$S_{mn} = mn\frac{T_1 + T_{mn}}{2} = mn\frac{\frac{1}{mn}+1}{2} = \frac{mn(mn+1)}{2mn} = \frac{mn+1}{2}$$
Q.E.D.
