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? 
 A: A symmetrical (but slightly unorthodox) approach:
Multiplying the $m$th and $n$th terms by $mn$:
$$mn\cdot \overbrace{T_m}^{\frac 1n}=m\\
mn\cdot \overbrace{T_n\;}^{\frac 1m}=n$$
As an AP is linear it only needs two points to be defined. Hence we conclude that 
$$mn\cdot T_r=r$$
Sum of the first $mn$ terms is
$$\frac {mn}2\big(T_1+T_{mn}\big)=\frac {1+mn}2\;\;\blacksquare$$

NB: It is clear from the above that 
$$T_r=\frac r{mn}$$
and 
$$T_1=a=d=\frac 1{mn}$$

See also this video here for an interesting analytic geometry approach.
A: 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,
$= \frac{mn}2 × \frac {1}{mn}\left[ 2 + (mn - 1) \right]$
$= \frac{1}2 × \left( mn + 1) \right)$
A: 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
$\sum\limits_{k=0}^{nm}a_k=\sum\limits_{k=0}^{nm}a_0+kp= \frac{1}{nm}\sum\limits_{k=0}^{nm}k= \frac{1}{nm}\frac{(nm+1)(0+nm)}{2} = \frac{1}{2}(nm+1)$
A: 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 $$\sum_{t=0}^{m\cdot n} u_t = \sum_{t=0}^{m\cdot n} \frac{t}{m \cdot n}$$
$$\sum_{t=0}^{m\cdot n} u_t = \frac{1}{m \cdot n} \sum_{t=0}^{m\cdot n} t$$
$$\sum_{t=0}^{m\cdot n} u_t = \frac{1}{m \cdot n} \frac{(m \cdot n)\cdot(m \cdot n +1)}{2}$$
$$\sum_{t=0}^{m\cdot n} u_t = \frac{m \cdot n +1}{2}$$
