Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

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.

share|improve this answer
add comment

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

share|improve this answer
add comment

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:


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


share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.