 Hello and welcome to another session on sequence and series and in this session. We are going to discuss Selection of terms in an AP. What does that mean? There are a few observations that You know few given terms are always in AP. So, you know, what are these terms we are going to discuss? So let's say when can we say that three terms are always in AP or four terms are always in AP or Five terms are always in AP. So is that is that some kind of or is there some kind of? structure of or let's say some kind of terms and In you know inherently they are in AP. So are there such terms? So that's what we are going to discuss today So let's take up Three first first the case of three terms. So let's study this three terms are always in AP three terms are always In AP, okay, let those three terms are a Or instead of A and B, let me use terms like P Q and R Okay, so PQ are are in AP. So do we have some, you know, generalized PQ are so that it will they will always be in AP, let's try and understand So when is P and Q and R in AP clearly when Q minus P the common difference must be equal to R minus R minus Q or Two Q must be equal to P plus R Okay, so do we have some selection of P and R and Q so that they are they are always in AP. So What we can do is let us say this is equal to or you know, so if you select P as let's say a minus D and Q as A and R as A Plus D where A and D are any arbitrary real numbers where where A and D are real numbers Are real numbers. So if you see A minus D A and A plus D will always be in AP, isn't it? Why because their common difference is always D Common difference is equal to D Let's take an example. Let's say A is 2 A is 2 and D is 3 Okay, A is 2 and D is 3. So first term is 2 minus 3 Second term is 2 itself and 2 plus 3. Clearly these three terms are in AP, isn't it? Minus 1, 2 and 5. What is the common difference? Common difference is 3 So we see that if you have A Minus D and A and A plus D three terms are there in this form They will always be in AP. So is it restricted only to two terms? No, let's go for what if there are four terms in AP Four terms always in AP Always in AP what all they can be. So A minus 3D A minus D A plus D and A plus 3D if you Look at them very carefully. The common difference is 2D right in these in these four terms common difference Common difference is equal to 2D Okay, so they are very much in AP with common difference 2D. So you can treat this as first term and Common difference to be 2D and then you can write any AP in this form. So all if an AP is in this form A minus 3D, A minus D, A plus D, A plus 3D Then they are in all they are always in AP. So hence Let's say there is a problem where you have to consider four terms in AP So you can always consider these like them instead of taking them as PQRS or A, B, C, D or A plus A, A plus D, A plus 2D, A plus 3D instead of that you can take A minus 3D, A minus D, A plus D and A plus 3D Okay, so example, let's say A is equal to 5 and Let's say B or D is equal to 1 then what are the terms 5 minus 3 5 minus 1 5 plus 1 and 5 plus 3. Is it it? If you check this is 2 This is 4. This is 6 and 8. They are in AP, isn't it? So hence Whatever be the values of A and D, they will always be in AP Again, if you have let's say you have to pick up You know 5 terms in AP, 5 terms 5 terms in AP. Okay, if you have to take up 5 terms in AP then clearly you see A minus 2D, A minus D, A, A plus D and A plus 2D with a CD of or with a common difference of D. Correct? Again, you can write 5 terms in AP like that. You can check on it, you know It's a common difference between every two consecutive term is D correct Okay, so What if we have a selection of 6 terms So if there is a selection of 6 terms then again A minus 5D A minus 3D A minus D, A plus D, A plus 2D, sorry 3D and A plus 5D So whenever there are even number of terms so we have You know A minus some odd multiple of D like that, right? So this is this is odd multiple of D, A minus odd multiple of D or A plus odd multiple of D These are 6 terms in AP all the time. So 6 terms in so these are 6 terms in AP always in AP. Similarly, there is no restriction for the next you have would have guessed that A minus 3D A minus 2D A minus D, A A plus D, A plus 2D and A plus 3D all are These are all 7 terms in AP Correct? There is a reason why we are writing in this manner. Why because let's say when we you know Take up problems where some of the terms of AP is given Then you can eliminate one variable that is D by adding all of them if you see if you add all of them Adding all of them just eliminates D and hence we have You know represented them in this format, right? adding all of them Eliminates D and if you know some of three terms or four terms or five terms or six terms or seven terms or for that matter any end term You can take terms in such a way that you know you eliminate D by adding So if you add all of them D is eliminated you get A directly and hence the reason why we are expressing them in this Fashion it will be clear when you solve problems on this subsequently