 Hello and welcome to another problem-solving session on sequences and series and we have been dealing with arithmetic progression problems and Specifically the nth problem nth term of an AP problems related to nth term, right in this question It said that pth and qth and arithms terms of an AP are AB and C respectively and You have to show that this term this expression is zero Okay, so I hope you understand what is pth term so p and q and r are the indices I did that is a position the term number in you know the location of that particular term correct and ABC are you know the the value of the term right? So it's like sixth term seventh term eighth term are 10 9 and 11 something like that So you have to prove this one. Anyways, so Assuming that you know an understood the question. So how do we solve it? Okay, so this is customary thing to you know start with a few, you know terms like let's say t1 is equal to the first term first term and D is equal to common difference, right the two vital characteristic of NAP these common difference, right? Okay, so now if that is so then we can say that a is equal to T1 plus p minus 1d Correct that is how nth term is found out. So b is equal to t1 again plus this time q minus 1d and C is equal to t1 plus r minus 1d So if you if you you know have managed to get these three equations, let's say 1 2 and 3 then Your job is kind of done Okay, now Let's try and find out q minus r. So yep. So or Yes, so you can start from this as well for example A minus B. So let's try to find out a minus B. So a minus B if you see Will be even and even will get cancelled and D and D also So let me just write so that it becomes easier. I will write for one so that it becomes easier for others as well So minus T1 minus qd plus D, right? So T1 disappears D disappears So it is nothing but D common p minus q p minus qd is a minus B Okay, so basically guys, this is You know, so we you can find out p minus q from here so p minus q is a minus B by D Is it it p minus q and now look at the given expression you need p minus q for sure But you need a C also Never mind. So we can get a C. How just multiply both sides both these sides by C so I can do this Is it it so what do I get? I get C C times p minus q is equal to AC minus BC upon D Right. This is let's say for now Let's do Two minus three. So this was simply one minus two. Let's do two minus three So two minus three will give you what? Again, so I'm not going to repeat the same process. So you'll get B minus C is equal to Q minus R times D just like here cyclical expression B minus C is Q minus Rd So Q minus R will be B minus C upon D and let's go back to the expression which we wanted we want A times Q minus R is it so we can multiply both sides by a By a without any problem. So A times Q minus R is a B minus AC by D. This is let's say five Okay Now here itself. Let's do Three minus one. So you would have guessed it by now. That's a cyclical Process so C minus a would fetch you R minus P times D So again from here, we can find out R minus P is equal to C minus a upon D And what term did we want? R minus P if you check B is here. So I want a B Attached attached with it. So multiply both sides by B. You will get B R minus P is equal to B times C minus a by D Fair enough this is six Right now look closely at four Five and six and add of all of them. So if you add you'll get C times P minus Q plus a times Q minus R plus Plus B times R minus P all LHS added together all RHS. Let's add We had a C minus BC upon D plus a B minus a C upon D Plus so this was how much this was if you see this is BC minus BA upon D So plus BC minus BA upon D and my friend if you see this Result D is a common denominator and let's write a C minus BC plus a B minus a C Plus BC minus a B Correct. So this BA I have written as AB here, right? And you see all of them are getting cancelled a B a B BC BC So it is eventually zero and this is what we wanted to prove in such cyclical You know expressions, what you need to do is simply find a Q minus you could have started with yeah, finding Q with respect to a and and P with respect to B and C with respect to R and all that but you know cyclical expression there's a difference to try subtracting and We the moment we got you know P minus Q B understood the hint we multiplied by the required remaining factor and we got an expression on This C times P minus Q and then if one was like that the other two have to be like that It's a psych no property of cyclic expressions and you figured out all the three separately added together and You got that is a result So one learning is that you need not be you know intimidated by a big expression Many a times where there wherever there is a cyclical expression, right? where you know this and This and this there are three parts where if you keep on replacing a by B B by C And keep on replacing P by Q and Q by R. You get the three terms such cyclical terms You can treat Independently and then you see a pattern and get the desired result. Okay. This is what the solution demanded here