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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?

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    The title of the question has been truncated. – 2017-02-10
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    Why the `\textrm`s all over the place? – 2017-02-11
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    "What do I do further?" Solve for $(a,d)$ the system $$a+(m-1)d=\frac1n\qquad a+(n-1)d=\frac1m$$ and deduce the value of $$\sum_{k=1}^{nm}a+(k-1)d=anm+\frac12nm(nm-1)d$$ – 2017-02-11
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    @ did, what is $a+(k-1)d$? – 2017-02-11
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    Sorry but what? – 2017-02-12

4 Answers 4

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

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.

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    @ hypergeometric, why is AP linear? – 2017-02-10
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    Because the difference between consecutive terms is the same. Similarly, for a straight line $y=mx+c$, the increase in $y$ is the same for consecutive unit increases in $x$, i.e. for $\Delta x=1$, we have $\Delta y=m$. – 2017-02-10
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    only needs two points to define the relationship.what does it mean? @ hypergeometric – 2017-02-10
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    In the straight line $y=mx+c$ the unknowns are $m,c$, which can be found given two points on the line. Similarly for the AP $a+(n-1)d$ where the unknowns are $a,d$. – 2017-02-10
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    @ hypergeometric, how does it happen ::?we conclude that – 2017-02-10
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    The conclusion can be drawn as it holds for two points (or two terms). If it holds for only one term, the same conclusion cannot be drawn. – 2017-02-10
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    @ hypergeometric, how did you get that ;$mn.T_r=r$ – 2017-02-10
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    Note that this solution is to illustrate the beauty of exploiting the symmetry of a problem to arrive at a nice short solution. However, for exam purposes it may be safer to use the standard plug-in-the-formula method, to earn full marks :) – 2017-02-10
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    Because the relationship has been shown to hold for $r=m$ and $r=n$. – 2017-02-10
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53411/discussion-between-s-ramanujan-and-hypergeometric). – 2017-02-10
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    @ hypergeometric, what is r? – 2017-02-10
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    $r$ is the general index of the AP, i.e. the $r$-th term, $T_r=a+(r-1)d$. Usually $n$ is used but here it has already been taken for a specific purpose. – 2017-02-10
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    what formula did you use for sum? – 2017-02-10
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    $\frac N2(a+\ell)$. That should be quite enough info for you to work on. – 2017-02-10
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    @ hypergeometric, how do you know that $T_1=1$? – 2017-02-10
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    @ hypergeometric, and also how is a=D? – 2017-02-10
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    See edits. Put $r=1$ into formula to get $a$. Put $r=2$ and subtract $a$ to get $d$. – 2017-02-11
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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,

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

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

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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

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

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    @ Jaroslaw Matlak, what is $a_k=..$? – 2017-02-10
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    @S.Ramanujan - it's formula for k-th element of arithmetic sequence – 2017-02-10
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    @ Jaroslaw Matlak, Then $a_0$ is the first term? – 2017-02-10
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    @S.Ramanujan Actually it's the 0-th term. First term is $a_0+p$ – 2017-02-10
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    @ Jaroslaw Matlak, Then what's $P$? – 2017-02-10
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    @S.Ramanujan $p$ is the sequence's common difference. – 2017-02-10
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    @ Jaroslaw Matlak, I didn't get completely, will you explain me here ;;http://chat.stackexchange.com/rooms/53289/keep-typing – 2017-02-10
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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 $$\sum_{t=0}^{m\cdot n} u_t = \sum_{t=0}^{m\cdot n} \frac{t}{m \cdot n}$$ $$\sum_{t=0}^{m\cdot n} u_t = \frac{1}{m \cdot n} \sum_{t=0}^{m\cdot n} t$$ $$\sum_{t=0}^{m\cdot n} u_t = \frac{1}{m \cdot n} \frac{(m \cdot n)\cdot(m \cdot n +1)}{2}$$ $$\sum_{t=0}^{m\cdot n} u_t = \frac{m \cdot n +1}{2}$$