 In this question it's given that in an AP, sum of m terms is equal to n and sum of n terms is equal to m, then we have to prove that sum of m plus n terms is minus m plus n, okay? So let's solve this problem. What is given? Let's first write that it's given sm. So first m terms is n and sn is equal to m and we have to find out sm plus n, right? So play with subscripts only, okay? So let's start sm. What is sm guys? Sum of first m terms of an AP is given by m by 2 twice a plus m minus 1d, a and d are not known. So and what is a? a is first term and d is common difference, okay? So what is sn then? So and sorry, this is equal to n, okay? Let it be 1 and n upon 2 2 a plus n minus 1d is equal to m, correct? Right? Now this is 2. Let's try and find out smn. So what is smn? Let's write it so that we have a track of what exactly we need m plus n by 2 and we need 2 a plus m plus n minus 1 times d. This is what we are looking for, okay? Okay, so how do we go about it? So either let's subtract these two equations and see what happens, okay? So subtraction will kill couple of variables. So, okay? Yeah, so if you see, let's first simplify this by multiplying the two equations by 2. So hence we'll get m 2 a plus m minus 1d is equal to 2n, correct? And n 2 a plus n minus 1d is equal to 2m 3 4. Now what? Let's subtract 3 and 4. So 3 minus 4 will give me what? So let's, you know, open the brackets and do the calculation. So the first term will be 2am plus m times m minus 1d and here minus 2am minus n times n minus 1d and here you'll get 2n minus m, correct? So if you take 2a common here, so 2a you'll get m minus n and from here let's open this again. So it is m square minus m minus n square plus nd, correct? So this is 2n minus m. Now what? So 2am minus n plus you can write this as m square minus n square minus m minus nd, correct? And this can be written as minus 2m minus n. So just change the sign on the right-hand side. So that I can swap the places of the terms inside the bracket, okay? Now this implies 2am minus n plus m square minus n square can be written as m minus n times m plus n and this is minus m minus nd, correct? Just check. Go slowly if you are not able to track it, correct? So m square, I simply wrote m square minus n square as this m minus nm plus n. Why? It will be clear in this very step. So now I can take m minus n common in this part. m minus n is common right now. So m minus n and within brackets m plus n and minus 1 times d and this is minus 2m minus n. Now you see you can cancel all m minus 1 because they are common to all, common factor to every term. So m minus n gets canceled assuming m is not equal to n because if it is 0, if m is not equal to n then m minus n becomes 0 and you can't cancel 0 from the entire equation. Okay? So what do we get? We get 2a plus m plus n minus 1d and this is equal to minus 2, isn't it? Now if I multiply the entire equation by m plus n by 2, right? So this will become 2a plus m plus n minus 1d and this one becomes minus 2m plus n by 2. Why did I multiply? Yep. So I will write multiplying, multiplying the equation by m plus n by 2. So when I did that, I get this equation. So the moment you get that equation, now you can see what is this LHS is nothing but s m plus n. This is the formula for s m plus n, isn't it? And what is there in the RHS? These two gets canceled so minus m plus n. So hence this is what we had to prove that sum of m plus n terms, see? Sum of m plus n terms which is s m plus n is minus m plus n and we got the same result. Hence proved, isn't it? What is the learning? So write the given formula, use the formula of sum of n terms of n, a, p and then we wrote two equations, manipulated, subtracted and got the desired result. So we keep track of what exactly you have to find out. So while we had this thing in our mind, we got this expression while doing the manipulations and then we multiplied simply by m plus n by 2 to get the desired result. Hence this problem was solved.