 Hello friends welcome again to a problem-solving session on sequence and series So far we have been discussing the general term of an arithmetic progression And in this question you can see it says that if the mth term of an AP B1 by n and nth term B1 upon m Then show that it's m nth term is 1 Now what is mn and mn and all that quite confusing, isn't it? But don't worry. Let's you know, let me first illustrate this problem with an example and Then it becomes very easy for you to solve. So what it means is mth term. Let's say m is equal to 5 Okay, so mth term will be fifth term Okay, so tm will be fifth term and let's say n is equal to 6. So it will be sixth term Right. So what will be tmn that is mnth term will be m into n That is 30 in this case Understood. So this is what it means. So don't get confused by these letters use of these letters Okay, now so once you understand the problem. So what they are saying is tm is equal to 1 upon n Right. So mth term is tm and that is equal to 1 upon n and tn is 1 upon m Okay, and you have to find out e mn. In fact, e mn is equal to 1 you have to show You have to show this. Okay. Let's try and solve this problem. So let us say let us say that The first term first term is equal to a and Common difference common difference is equal to D Okay, then tm will be One upon n is given as tm. This can be equated to first term plus m minus 1 b Is it this is equation number one similarly 1 upon m is equal to first term plus n minus 1 d Right because this was mth term and the second is nth term Okay, guys. So what we can do is do this operation one minus To okay one minus two what will you get you'll get one minus n minus one upon m in the LHS and here a plus md Minus d was the first one and then subtract the second one minus a Then minus n d and plus d, right? This is what it will be. So this d and this d goes this a and this a goes So what do we see? We see that and we can simplify this as well. So, right and this is equal to um m minus n d correct so Assuming m is not equal to n right assuming m is not equal to n then we can cancel this If m is equal to n then we can't cancel this. Please keep it in mind. So assuming right m is not equal to n, right? So hence d. What do I get 1 by 1 by mn? So we found out d. Let's find out a so From here we can find out a a will be equal to 1 by n Minus m minus 1d and d I can write as mn which is equal to 1 by n minus 1 by n Again, why because this m into 1 by mn will give you or let me write it so that you all will understand clearly So 1 by n minus m into 1 by mn plus 1 upon mn So this and this goes so it is 1 upon mn as well. So a also happens to be 1 upon mn Okay, so hence we had to find out t mn. So what will be t mn guys t mn is nothing but first term plus mn minus mn minus 1d right, this is the Result so a is 1 upon mn and then this is mn Minus 1 times 1 upon m So hence if you see this is nothing but 1 upon mn Plus mn into 1 upon mn. Let's open the bracket and this is 1 upon mn, right? so mn times mn is 1 and this thing and this thing will go so hence this is simply 1 and This is what we needed to prove is it see they were asking you to prove that mn's term is 1 and we have proved it here Okay, so I hope you understood the process. I Just used only one concept and that is the concept of finding the nth term of an AP