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 $(mn+1)/2$

Question

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?

Answer 1

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 $$\frac {mn}2\big(T_1+T_{mn}\big)=\frac {1+mn}2\;\;\blacksquare$$

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

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See also this video here for an interesting analytic geometry approach.

Answer 2

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,

$= \frac{mn}2 × \frac {1}{mn}\left[ 2 + (mn - 1) \right]$

$= \frac{1}2 × \left( mn + 1) \right)$

Answer 3

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

$\sum\limits_{k=0}^{nm}a_k=\sum\limits_{k=0}^{nm}a_0+kp= \frac{1}{nm}\sum\limits_{k=0}^{nm}k= \frac{1}{nm}\frac{(nm+1)(0+nm)}{2} = \frac{1}{2}(nm+1)$

Answer 4

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 $$\sum_{t=0}^{m\cdot n} u_t = \sum_{t=0}^{m\cdot n} \frac{t}{m \cdot n}$$ $$\sum_{t=0}^{m\cdot n} u_t = \frac{1}{m \cdot n} \sum_{t=0}^{m\cdot n} t$$ $$\sum_{t=0}^{m\cdot n} u_t = \frac{1}{m \cdot n} \frac{(m \cdot n)\cdot(m \cdot n +1)}{2}$$ $$\sum_{t=0}^{m\cdot n} u_t = \frac{m \cdot n +1}{2}$$