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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

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3 Answers

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.

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then you have to count the sum of $T_1,...,T_{mn}$ it seems, by computing $S_{mn} = \frac{T_1 + T_{mn}}{2}mn$...

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If $\,a_1,a_2,....\;$ is an arithmetic progression with common difference $\,d\,$ , we have that

$$S_r:=a_1+a_2+\ldots +a_r=\frac r2\left(2a_1+(r-1)d\right)$$

In your case, and using what you already did:

$$\frac{m+n}{mn}=\frac1n+\frac1m=2a_1+(m+n-2)d=2a_1+(m+n-2)\frac1{mn}\implies$$

$$2a_1=\frac2{mn}\implies \color{red}{a_1=\frac1{mn}}\;\;,\;\;\text{and since also}\;\;\color{red}{d=\frac1{mn}}\implies$$

$$S_{mn}=\frac{mn}2\left(2a_1+(mn-1)d\right)=\frac{mn}2\left(\frac2{mn}+1-\frac1{mn}\right)=\frac12+\frac{mn}2=\frac{mn+1}2$$

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