 So I'll be starting with the chapter sequences and series by the way this chapter is also known by an alternate name progressions and series progressions and series okay now why two names or what is the difference or similarity between these names let's talk about that first before we start solving anything. So since you have done this chapter I think I'm assuming everybody would have touched up on this chapter in school. I would first like to know what is the difference between sequences progression or progressions and series what are the difference between these three terms are they the same of course I don't think so they should be the same because if they are same why would we you know give them three different names okay so who will tell me what is the critical difference between these three terms anybody who has done this chapter in school or maybe would like to take a try what are the difference between these three terms anybody let it be wrong no worries take a call just take a guess anybody Karthik, Mayur, Nidhi, Nivesh, Satya, Shalini, Prateesh, Tejaswini oh before I move on I would like to remind you that we have not completed the limits chapter we have only done it to the extent which is sufficient enough for you to do well in your school or semester exams okay we'll be continuing again with limits chapter once we are done with your first term exam I think everybody is writing their first term exam and this chapter that we have taken up it's a part of your first syllabus and that's why we are covering it up we'll come back to these chapters in much much more detail don't worry about it okay yes Prateesh you are starting a new chapter okay so sequence is basically a listing a listing of numbers it could be listing of numbers it could be a listing of alphabets it could be listing of words okay or it could be listing of any set of elements where there is a pattern where there is a pattern or a rule of this listing where there is a pattern or a rule governing the listing okay to give you some examples of to give you some examples of sequences let's say okay I'll ask you to tell me what kind of a sequence is this two three five seven eleven thirteen seventeen nineteen twenty three can anybody tell me what kind of a sequence is this yes exactly Karthik so it's basically a sequence of prime numbers it's a sequence of prime numbers right so when I was writing down this listing you could actually observe some rule or a pattern and on the basis of that you could basically guess that sir has written a sequence of prime numbers right so what are the pattern here every number that I have listed over here is either divisible by one or itself right that's the characteristic of every prime number we all know that okay let me write one more maybe you'll be able to recognize this as well one one two three five eight thirteen twenty one right so it's a Fibonacci sequence it's a Fibonacci sequence Fibonacci sequence okay so this is a sequence where you can see a pattern right what is the progression on the other hand anybody there are there is a full I think you can see a video on YouTube which is based on Fibonacci sequence so a lot of things are based on the Fibonacci sequence okay they relate it to music they relate it to arrangement of petals on a flower they arrange it to the structure of our ear so many things are really you just have to type Fibonacci on the YouTube video and they'll show you the photo of videos on Fibonacci okay anyways we'll talk about them in our you know offline session sometime okay now what is the progression progression on the other hand is a sequence is a sequence where the value of the term present on the n-th position or you can say the n-th term of that particular sequence is a function of its position okay so please read this as if you have the position of the term known then you could actually generate the value of that term in terms of its position that means the value of a sequence expression term is dependent on its position value right at which position it is located many examples you would have already seen in school something like 1, 3, 5, 7, 9 I'm sure you know what is this this is a straightforward arithmetic progression right okay arithmetic progression in arithmetic progression we know that I could write down my in this case particularly I could write down my n-th term as 2n-1 where n is the position of the term for example if I want the second term I would put n is 2 I will get a 3 straight away from this formula if I want the fifth term I will put 5 in place of n I will get 9 straight away from this formula okay so by the way if I go back if I go back to the sequence example which I gave you is the sequence of fine number of progression can I call this sequence as a progression by the way remember progressions are sequences only so sequence is a super set you can say that and progressions are nothing but they are special sequences where you have a formula for the value of a term in that sequence given its position is known so is there any formula by which if I tell you the position of the prime number putting that position in that formula you can get the value of that prime number there is no such formula by the way right mathematicians are still you know breaking their head to find a pattern but prime numbers prime numbers I can say the randomness of the prime number is as good as the randomness of the digits of pi okay so there is no concrete formula that we are aware of where you could where you could put the position of a term okay or you could put the position of a prime number let's say I want 766 prime number and you put it there and it throws out the 766 prime number there is no such formula unfortunately right but is there any formula for Fibonacci sequence can this sequence be called as a progression can I call this as a Fibonacci progression I'm asking you a general knowledge question by the way it's not a question related to our CBSC or J curriculum exactly Karthik there is a formula okay and that formula is called the Binitz formula okay so please do us some research work on it so there is a formula by which you can find out we can find out the value of a Fibonacci sequence term given its position is known by the use of Binitz formula okay Binitz formula okay just do a quick read up on this it is available on Wikipedia or any any kind of a YouTube it basically comes from a kind of a relationship between the three terms so there is something related to characteristic equation and all no need to go into very depth of it just to give you an idea that Fibonacci sequence is actually a progression okay another example of a progression that I could cite here is something like you know a one three nine twenty seven eighty one right what is this by the way everybody knows that this is a geometric progression this is a geometric progression okay here what is your nth term if you want to write down your nth term what is it is 3 to the power of n minus 1 right if you put 1 you get 3 to the power 0 which is your first term if you want 3 you have to put n as 1 you get 3 n as 2 which you which will give you 3 to the power 1 which is 3 okay so any term let's say I want the fifth term I have to put n as 5 over here so I'll get 3 to the power 4 3 to the power 4 is 81 okay so there are many there are many such you know progressions that we could write on I'm just you know stopping it over here but the idea is to let you know that progressions are nothing but sequences where you could find out the value of a term in that sequence given its position is known note that all sequences are not progressions but all progressions are sequences it's like all alkalis are bases but all bases are not alkalis right I'm giving some chemistry example also yeah so what is series on the other hand anybody series on the other hand okay so series of the other hand is nothing but if you add the terms terms of seek of a sequence added okay sequence added or subtracted I mean subtraction is the kind of an addition with the negative of that term it gives you a series okay please do not interchangeably use a sequence and series you know words with each other there was a student who I asked what is this one one two three five immediately he said said this is Fibonacci series it is not a Fibonacci series unless until you add them it's a Fibonacci sequence right tomorrow you may be attending a kbpy interview and you say a sequence to be series the professors are going to give you a hard look at what he's saying okay so please be careful choose your words carefully leonidus okay is it fine I hope you have anybody who's heard of this dialogue which I just now give choose your words carefully leonidus they may be your last as king no what if people don't watch movies also anyways so what are we going to study in this chapter we are going to study in this chapter the following things but I may not be covering all the things before your semester exams I may be doing a few of the things after your semester exams so overview of this chapter we are going to talk about primarily three types of progressions okay what are they arithmetic progression okay I'm going to cover this today we are going to talk about geometric progression this also I'm going to do today geometric progression harmonic progression this I'm not going to do today why because I'm sure in schools they have not taken up harmonic progressions okay so we'll be not talking about it we'll talking we'll be talking about it once your semester exams is done with okay but still if you have any questions any concerns related to anything which I'm going to specify here please highlight okay then we're going to talk about means not the means you have studied in statistics we are going to talk about arithmetic mean we are going to talk about geometric mean we are actually going to talk about harmonic mean after your semester exams okay and also we'll talk about inequality inequality among means among the means okay so AM GM HM inequality in fact today I'll talk about just AM GM inequality post this we are going to talk about series okay series is another subtopic that we need to talk about in series we are going to first talk about special series okay special series are basically incorporates some of natural numbers some of squares of natural numbers some of cubes of natural numbers like that why we why we call them special series is because we use those series to basically find out summation of complicated series okay so they they serve as the basis for solving complicated kind of a series so we'll be talking about that but not in today's class because I'm sure in school series has not been taken up am I right am I right anybody can confirm that Nikita Shalini not taken right see Karthik has yeah it's not taken right okay so we'll again talk about it because without series this chapter is just going to be like you know you know half half of the size okay so we have to talk about series I have no clue why it was not taken in school maybe they have deleted it from the the current syllabus okay so special series we are going to talk about we're also going to talk about the methods of summation methods of summation okay and that would include that would include sigma method of summation sigma method of summation see I'm just giving you an overview of the entire concept maybe I will not be covering everything in today's session okay we'll be talking about methods of difference method of difference method of difference including Newton's difference method and we'll be talking about one very interesting you know part of method of difference but I normally write it as a separate subtopic because it takes considerable amount of our time and that is called the vn method okay vn method we'll talk about it don't worry as of now these are the terms which have you have not heard of don't be surprised okay in series I'm going to talk about one more thing AGS which is called Arthmetico geometric series okay so these are the topics which I'm going to touch upon in this entire chapter but for the purpose of school exams I'm going to touch upon the following this one this one this one this one and this one okay so today our agenda for the entire class would be covering these topics okay so are you all ready I mean you have already done arithmetic progression in class 10th haven't you anybody who has never done arithmetic progression before something little bit you would have done I'm sure okay don't worry I'm going to start everything with scratch I'm not I'm not starting with any pre assumption that you already know it so I should not skip it or something like that okay so let's start let's start our discussion with arithmetic progression what is an arithmetic progression arithmetic progression is a sequence right where the listing are such that every subsequent term differs from the preceding by a fixed value and that fixed value is called the common difference so it is a kind of a listing or it is a kind of a sequence where where every subsequent term I'm sorry I wrote a 2d again my bad okay dot dot dot every subsequent term that means let's say every tk plus one next term minus tkth term this difference is a constant and this is called the common difference this is called the common difference okay now many people call ask me sir is this supposed to be always a non-zero value no it could be zero also it could be negative also it could be positive also and to your surprise it could be imaginary also yes it could also have a common difference which is imaginary okay but primarily our discussion will revolve around such arithmetic progressions where the common difference will be real so we call them as real arithmetic progressions okay so we're talking about a real arithmetic progression but yes don't be surprised if you see a anomaly if you see a diversion and some question comes abruptly in your competitive exam where the common difference is not real so don't be afraid the concept the basic principles will be the same it will not going to differ okay now having known this characteristic of this particular sequence which makes it an arithmetic progression can I use this to find out any nth term of that progression because progressions are known to uh basis sequences where we can find out the value of a term in a sequence given its position is gone so can we use this characteristic of this particular sequence to get the nth term I know you have already done in class 10 you will immediately say sir a plus n minus 1 b right but can we obtain it from here very simple all of you please pay attention let's start putting k value as a one if you put k value as a one it becomes t2 minus t1 as a d put k value as a two it will become t3 minus t2 as a d correct yes or no I'm just I'm just changing this k over here put k value as a three it gives you t4 minus t3 as a d okay and just keep on putting these values till we reach let's say n minus one so if I put k as n minus one I'll get tn minus tn minus one is equal to d right let's add these terms let's add these terms now while I'm adding the terms on the left hand side you'll see a very interesting thing happening these terms will start cancelling out okay the term before this will cancel out this of course I have not written everything down in the interest of time so what will survive while I'm adding terms on the left hand side will only be tn and t1 so tn and minus t1 will survive let's all the terms will get cancelled off correct and here I will add d d d d d d d d d how many dds are there tell me how many ds will be there anybody anybody n minus one d is right right right so from here can I say tn is t1 plus n minus one d right and since you have assumed your t1 to be a you can safely say that your nth term of an arithmetic progression is this many people say sir I use so much of hard work you did to get this you could have easily observed the pattern here of course yes you can also observe the pattern I did it more mathematically okay so this is the nth term of a arithmetic progression please note this down correct Karthik Sharduli okay so on the basis of this can we start with a small question okay so let's start with a very very small question just on this formula only before that please copy things down if you have any questions do let me know done Karthik good okay so let's take a question let's take a question okay so I have a question here consider two arithmetic progressions s1 and s2 okay so read them as sequences so s1 sequence is two seven twelve seventeen da da da da da there are five five hundred terms in that sequence s2 sequence is one eight fifteen twenty two da da da da 300 terms find the number of common terms that is first part of the question so how many terms would be common to both these sequences s1 and s2 and the second part of the question is find the last common term let me allow you some time to work on it so I think two two and a half minutes let's allow everybody to work on it those who are done I would request them to give me their response as to how many terms are common and please mention the value of the last common term okay okay so Arnav is given his answer very good Arnav okay so the trick here is everybody please pay attention we need to identify the first term which is common to both of them okay and then from there on I will tell you a trick by which you can get the subsequent terms which are common noel excellent very good so noel Arnav has given their answers well done okay so if you list if you see the see the listing of the sequence s1 okay I'll just write few more terms uh so I think the common difference is a five so this will be 22 27 32 da da da da okay so this is a sequence whose common difference let me call d1 as is a five okay s2 sequence s2 sequence s2 sequence s2 sequence is 1 8 15 and okay I don't need to go any further because I figured out 22 is occurring in both of them yes so this is the first thing that we needed to do we wanted to find out which is the first common term occurring between them okay and of course as you can see this is a sequence where the common difference is a seven okay by the way common difference if somebody uses this word common difference it is always the difference of succeeding minus preceding not the preceding minus succeeding are you getting my point common difference is always the term which is coming later minus the term which is coming before that okay not the other way round okay now so let us say I start listing down the common terms over here so common terms the first term will be first term will be a 22 can somebody tell me what would be the next term which will be common can anyway tell me simple logic I want to just see your simple logic over here after what difference will my second term appear ah good try karthik noel absolutely right the second term will appear after a difference which is lcm of d1 and d2 okay so what is the lcm of 5 and 7 lcm of 5 and 7 is 35 so the next term that will come will come after a difference of 35 that means the next term will be 57 okay 22 plus 35 next term will be 92 now why why do you see it's a very simple thing let's say in your school you have two bells one bell rings after every five hours okay one bell rings let's say five minutes five hours is too much let's say one bell rings after every I mean a gap of every five minutes other bell the second bell rings after a gap of seven minutes let's say both of them ring right now then when will be the next time when they will ring together after lcm of 5 and 7 which is 35 minutes okay same thing is happening in this scenario also so you can treat sequence s1 as the time duration when the bells are ringing right so s1 and s2 they're ringing at a time gap of five minutes and seven minutes so if they have run now at 22 then they will ring again after a time gap of 35 minutes which is the lcm of 5 and 7 okay now till where I have to go how will I know that now you will say sir very obviously you have to go till the lesser of the 500th term or the 300 term whichever is the lesser of the two you can only go to that extent isn't it because our common term will be decided by which is the limiting reagent again I'm using chemistry term okay the limiting reagent is the one which is having a lesser value right so first we need to find out what is the 500th term in s1 okay that is very simple we'll use the formula just now we saw so a plus n minus 1 into d so this is I think 2497 right and what is the 500th term in s2 so that will be I believe 300 minus 1 into 7 right so this is 2100 minus 6 which is 2094 am I right okay so now this term will go to such an extent where this term has to be lesser than or equal to 2094 because out of the two the lesser is 2094 so I can go only till 2094 or before that I can't definitely surpass it okay so here I would use my formula a plus n is not known I don't know how many terms are common okay so a plus n minus 1 into d is equal or less is less than equal to please don't put equal to have to take it as less than equal to now remember here that n is a natural number so we have to find a natural number the greatest natural number which satisfies this criteria right the greatest natural number which satisfies this criteria that will be my number of common terms okay so let's try to solve it so this is going to be 2072 I think if you divide by 35 you will get 59.2 ish so n has to be less than 60.2 so the greatest natural number lesser than equal to 60.2 is your 60 that means there are 60 terms common to both the sequences there are 60 terms common to s1 and s2 okay so this solves the first part of the question second part of the question is how many what is the last common term yeah last common term will be a 60th term so 60th term will be n plus n minus 1 let me write it completely 60 minus 1 into d35 how much does it come out to be it should come out to be 2087 okay so the answer to this question is 60 is the last you know common term or they have all together 60 common terms where the last term common is 2087 is it fine any questions any concerns anybody okay so without much space of time we'll now talk about we'll now talk about the sum of n terms of an arithmetic progression okay this also is very much a familiar concept to all of you sum of n terms terms of an AP but before I give you the formula or before we derive the formula for n terms of an AP I would like to share with you a very small story which happened almost 300 years ago in a school in Germany okay where there was a school teacher she was teaching a grade of you can say class 1 or class 2 grade okay they were children of the age group of 10 years 9 to 10 years okay and this teacher gave these students a question in fact she wanted them to be engaged so they she gave them to add all natural numbers from 1 to 100 okay I write enough you already know this story so this German teacher she gave the students to add natural numbers from 1 to 100 thinking that this will keep all the students engaged right of course she was not expecting a out of you know turn answer coming from somebody so quickly but there was a small child who basically you know solved this question much earlier than what the teacher expected it to be so he did one thing he took the same series and he wrote it in a reverse fashion okay something like this right and if you write these numbers in a reverse fashion obviously it doesn't change the sum isn't it so what he did next was he added the two series so s plus s is a 2s and he added the terms one above the other or one below the other so one plus 100 101 then he added 99 and 2 which is again 101 then he added 98 and 3 which is again 101 then he added 97 and 4 which is again 101 and he kept on doing it and he realized that all these terms are giving you 101 101 101 right now that made his life easy because on the right side he had obtained 101 written 100 times isn't it so it's as good as 101 into 100 so s becomes 101 into 100 by 2 which is nothing but 5050 and he got the result pretty fast as compared to his his school you know friends this guy later on went on to become one of the greatest mathematician and physicist and his name was Carl Frederick Goss Carl Frederick Goss okay if anybody has prepared for a PRMO or RMO or INMO you would have definitely learned Lumber theory number theory and in number theory Goss has made subsequent amount of contribution he has made contribution to the field of integral calculus he has made contribution to the field of physics and electrostatics I'm sure in classroom the first chapter you'll be dealing is Goss law when you're we're talking about electrostatic chapter so this this gem of a mathematician was you know spotted much early in his life and he was just nine to ten years old when he solved this kind of a problem now we'll be using the similar approach what Goss used to find the sum of any generic or any general arithmetic progression so let's say I want to find out the sum of a a plus d a plus 2d a plus 3d and so on till let's say last term is nth term is a plus n minus 1d so we can write the same series in a reverse fashion and that would not change our sum okay so I'm just writing the same stuff backwards okay dot dot dot till the last term here will be a okay now add the two series like the way Goss did so this will be 2sn now add these terms the term which is you know in the same vertical line so this will be 2a plus n minus 1d okay add these two terms also this is also 2a plus n minus 1d okay add these two terms also this will also be 2a plus n minus 1d and like that you would realize that even till the last term it will be 2a plus n minus 1d so essentially what have you done you have written 2a plus n minus 1d n number of times and there you go your sum till n terms will become n by 2 2a plus n minus 1d okay now we already took the idea from Goss but we'll also do some analysis on this formula we will not just remember this formula in a dry fashion we'll do some a bit of analysis but first note down note down the result and note down the derivation now a couple of things that you would like to know note down over here number one the first thing that you should note down is you realize that while you were deriving this result you realize that the sum of terms equidistant from the beginning and the end was a fixed value in this case it was 2a plus n minus 1d isn't it see what have you done here a is the first term a plus n minus 1d is the last term so first and the last sum is 2a plus n minus 1d second and the second last term sum is also 2a plus n minus 1d third and the third last is also 2a plus n minus 1d and so on what does it tell you it tells you that this is a characteristic of any arithmetic progression where in any arithmetic progression the sum of right down in any ap the sum of the sum of terms equidistant equidistant from the beginning and the end is equal to the sum of the first and the last okay so here you realize that p1 plus tn is same as t2 plus tn minus 1 will be same as i'm so sorry i wanted to write it equal to yeah is equal to sum of third and third last term is same as why am i writing plus yeah the same as fourth and fourth last term and so on dot dot dot it is same as sum of some kth and kth last term okay so this is something which is a very very interesting property worth noting in any arithmetic progression and if there is a middle term it will be equal to twice of the middle term okay so it will be twice of the middle term if at all there is a middle term okay please remember middle term might not be always there okay if if it exists if it exists okay so if there is a middle term it will be twice of the middle term okay yeah that's my way of saying that okay so i'll just give you an example 1 3 5 7 9 okay let's let's take one more maybe 11 okay so 1 plus 11 is same as 9 plus 3 is same as 7 plus 5 okay so some of the terms equidistant from the beginning and the end will be will be fixed and that value will be the sum of the first and the last term so don't worry rameh i'll be i'll be using it you know many a times okay is it fine any questions okay we'll take one more where there's a middle term 1 3 5 7 9 11 and 13 so this was this 14 this was this 14 this was this also 14 and this is double of that 7 7 into 2 14 is it fine any questions okay correct absolutely right ramesh okay so this is the first thing that you should all observe about any AP any arithmetic progression this is going to be true okay all right next important property that you should know number number two property you would always realize that the sum of n terms of any arithmetic progression will always be a quadratic in n without a constant it's a quadratic but without a constant if you see a constant in any sequence please note that that sequence cannot be an arithmetic progression so if the sum of terms of a sequence or you can say sum of term of a series is a quadratic in n don't don't jump for the fact that it will be an AP no if that quadratic in n doesn't have a constant then only it will be an AP is it fine any questions okay now i'll give you a simple example to you know look into this see let's say i ask you this question and please let me tell you this observation many people do not make because they just you know remember the result but if they forget to analyze that result so let's say somebody gives you a question that hey there is a sequence whose sum of n terms is n square plus n plus 1 can the sequence be an arithmetic progression what will your answer be the answer should be immediately no illa sir why because one is there you don't have to try it out even i mean i will try it out for you and show you but don't waste your precious time of the exam to figure that out if at all you see a constant appearing in the sum it should be a quadratic in n but in that quadratic if there is a constant non-zero constant appearing it cannot be the sum of an AP for sure let's check it out so let's say t1 t1 is s1 correct the sum of one term is your first term only so t1 is 3 what will be t2 what will be t2 t2 will be s2 minus s1 s2 will be what if i'm not wrong four plus two six six plus one seven seven minus three s2 will be four okay what will be t3 t3 will be s3 minus s2 okay and if you put three you'll get 9 plus 3 12 12 plus 1 12 plus 1 13 13 minus 7 you get a six what did you see here is that there is a jump of plus one over here but there is a jump of plus two here can this happen in an AP you'll say no sir it cannot happen in an AP the common difference is fixed so it cannot be an AP it's not an AP okay so you don't have to do all these things in your examination you know hall and examination room to figure it out whether it was actually an AP or it was just some just some kind of a i can say china product okay so deceive you okay is it fine any questions okay so based on this i would give you a small question to solve let's take a small question so how are your exams going on charlie how are your exams going on how was the paper so far good easy okay okay why it should be easy no okay karthik uh fine only okay i know karthik you have one paper tomorrow and hats off to the people who have come for today's class despite having an exam tomorrow shows your commitment keep it up what about others shalini finnav i want to hear so i want to hear from nps hsr groups anybody and these do keep sharing your papers on the group or even you know you can share it personally with the faculty members okay okay pratej harshita mayur how are your exams going on nevesh i hope you're finding things easy with respect to uh uh sentence stuff right school stuff school stuff will be now uh would be appearing easier to you isn't it not yet started okay you know good to know that okay let's go on for a question so i have a question for all of you based on our analysis of the sum of n terms of an AP so here's the question read this question carefully the question says find the sum of first 24 terms of the AP a1 a2 a3 dadadadada ramesh dadadadada for you till 80 a24 if it is known that a1 plus a5 plus a10 plus a15 plus a20 plus a24 now please watch out for the subscripts over here they carry the information uh you know which is required for you to solve the question so if this sum is 225 what is the sum ah ramesh that was too early you could have just waited for a few more minutes absolutely right but you could have waited for a few more minutes please give me a response on the chat box everybody very good enough that was also quick and let me also take the attendance ramesh he is he is the best i'm sure he must be a physics lover sharduli okay okay shall we discuss it let's discuss it see uh if you look at these subscripts a1 a24 basically it is pointing towards the first and the last term of that progression right isn't it because we are summing till 24 terms so a24 supposed to be the last term a1 is supposed to be the first term a5 a25 is the fifth and the fifth last term and what did i say a couple of minutes ago that this sum will be same as this okay remember some of terms equidistant from the beginning and the end would be same as the sum of the first and the last term and twice of the middle term if at all the middle term exists okay and not only that a10 and a15 these are terms which are again 10th term and 10th term from the beginning and the 10th term from the end so their sum should also be same as a1 plus a24 correct so all these three sums will be same let's call them as xxx so xxx three times adds adds to give you 225 so 3x is 225 so x is 75 right now we can use the formula that sum till 24 terms is 24 by 2 sum of the first and the last term you could use any two terms which are equidistant fifth and the fifth last 10th and the 10th last doesn't make a difference because all these values will be the same so the answer will be 12 times 25 which is 900 and I think Ramesh was the first one to answer this correctly Vendal Naresh I think Arnab also got it right okay any questions anybody any questions any concerns any kasta in life okay no kasta in life then we'll move on to one more question okay okay let's take this question options are already given to you here let's read the question first let a1 a2 a3 till a4001 is an ap says that reciprocal of the product of a1 a2 plus reciprocal of the product of a2 a3 and so on till reciprocal of the product of a4000 and 4001 is 10 and at the same time it has also been given that a2 plus a4000 please make this as 4000 there's a small typo here okay so a2 plus a4000 is 50 then you have to find out the modulus of the absolute value of the difference between a1 and a4001 a1 and a4001 your options are already there on your on the screen 20 30 40 and none of these I'll put the poll on once you're done please you can either answer to me on the chat or you can press on your response on the poll as well wow I'm getting the responses already very good where school are you from Ramesh? Okay good good good well done this is a good question please make use of all the information given to in the question okay okay Ramesh I think we'll keep an upper cap of two and a half minutes or three minutes on this first that we can start the discussion okay last 30 seconds if you want to give your response please do so okay let the countdown begin five four three two one go okay so those who want to take a chance also please put your response so that we can start the discussion as of now I can see only 11 of you actively voting okay so let me end the poll the poll result shows that most of you have gone with option number B okay B for Bangaluru let's check whether B was right or not okay now see the first thing I need to you know notice here is that I have already been given the information about this okay now what is this actually A2 plus A4000 is as good as A1 plus A4000 one do you all agree with me on that that means if this is given means indirectly this is given right now looking at this expression looking at this expression I have a rough you can say roadmap in my mind what is that rough roadmap the roadmap is if I somehow can write this as okay correct me if I'm wrong under root of A1 plus A4000 one square minus four A1 A4000 one my job will be done and this information is already known to us as a 50 okay so basically what I'm doing I'm using the formula mod A minus B is what in fact if I say A minus B square is what it is A plus B square minus four AB isn't it so mod A minus B square will be under root of this okay so this is exactly what I plan to use right because I have already obtained A1 plus A4000 one from the given fact in the question okay so this has already been stated so my whole and soul ambition is now to get this guy first of all can I find this fellow A1 into A4000 one if I am able to find this town my life will be easier and I would be able to crack this question okay so this is the roadmap that I would have in my mind while I am trying to solve this question see note that when you go with the roadmap in your mind be prepared to fail also right it's not that I am a teacher and every time I make a roadmap to solve any question let's say I'm trying to solve some you know international olympiad question I okay let me try it like that when I'll be successful only there is no guarantee like this so when I start with the plan the you know I keep telling myself I may I may you know fall also sometime I may be proven wrong also sometime or maybe I may be you know not proceeding any further in that question so may I have to go back and take a recourse so always this is you can say a guru mantra that I would like to give you when you have a plan in mind don't be like okay this has to be successful no it never happens life is not like that right so you may fail also if it fails you have to come back and you know attempt once more or you know go from a different angle altogether okay so let's try to take this as our roadmap and I may fail also with that I will be proceeding now let's see what I can do I've not used the first information as of now so I'll just use that a 1 by a 1 a 2 1 by a 2 a 3 1 by a 3 a 4 okay and 1 by a 4000 a 4001 so this is given to us as a 10 okay now let us multiply throughout with a common difference d the common difference which is the common difference of this ap itself okay so where d is the common difference where d is the common difference please note that common difference could be a 2 minus a 1 also it could be a 3 minus a 2 also it could be a 4 minus a 3 also it could be a 4001 minus a 4000 also please note that the common difference doesn't change okay now I will be using this idea in this particular expression so this first d that you see on your very first term I will write that d as a 2 minus a 1 okay the second d that you see over here I will write that d as a 3 minus a 2 okay similarly this third d that you see I'll be writing that as a 4 minus a 3 okay and the last d that you see over here that d I will write it as a 4001 minus a 4000 and what is the benefit of doing that you must be thinking sir why are you doing this what will it help us okay let's see if I divide individually both these terms by the denominator can I say I'll get 1 by a 1 minus 1 by a 2 similarly here it'll give me 1 by a 2 minus 1 by a 3 similarly here it'll give me 1 by a 3 minus 1 by a 4 and the last one will give me 1 by a 4000 minus 1 by a 4001 now see miraculously many terms will start getting cancelled and it'll cancel to such an extent that you'll be only left with 1 by a 1 and 1 by a 4001 okay that is 10d let's take the lcm over here that will be a 1 a 4001 this will be a 4001 minus a 1 is equal to 10d so far so good everybody's happy till this step if no please highlight where you are not happy I will try to make you happy this is not an easy question by the way right this is of slightly j main caliber okay now a 4001 is nothing but a 1 plus 4000d as per our very first formula minus a 1 by a 4001 this is equal to 10d are now minus 4001 and plus 1 oh okay you got your mistake okay fine a 1 a 1 gets cancelled and not only that d d will get cancelled and one of the zeros one of the zeros will get cancelled so from here can I say 400 will be equal to a 1 a 4001 and thankfully this is what I needed and I got it okay so thank the god we have got this and I'm going to substitute it over here so my answer is going to be under root of 50 square minus 4 into 400 okay so this was 400 right so a 1 into a 4001 a 1 into a 4001 is 400 so I've just used this formula and my answer will come out to be 2500 minus 1600 that's nothing but under root of 900 which is plus 30 okay so which is option number b as most of you actually marked it yes absolutely right 30 is the right answer is this fine any questions any concerns anywhere please highlight clear all good I think final result is here if you want to copy this down also you can do so ah that's that's a question which you will get in every stages of your you know preparation day just any how do I suppose to know I have to solve the question in this way okay so there is there are some things which you will learn later later on also so this kind of a expression is basically this kind of an expression is basically seen in vn method of summation which probably will come across a little while you know after we are done with our semester exams whenever I'm discussing this chapter once again so literally down the chapter you will learn vn method where we write down a term as a difference of two terms okay and to create a difference of two terms I needed a 1 a 2 either as a difference a 1 minus a 2 or a 2 minus a 1 on the top and that gave me an idea to do it so there has to be a prior experience a bit but yes with more and more problem solving you will start getting those experiences don't worry about it it's a part of your learning now that you know this could be done so this is now there in your armory right this is a weapon which you can use in some other question okay so in your in your preparation time always be like you know from wherever you're getting the knowledge acquire it and see where you can apply it nobody is a born mathematician of course few of them are who are you know the prodigies but many of us we acquire knowledge while we are solving more and more questions that's why if you keep saying or do assignments solve from your centre module do all the supplementary books so as you solve questions you will automatically get these kind of you know approaches don't worry about it sure I can this part Karthik let me know once you're done great so now we are going to talk about now we are going to talk about the geometric progression because I think we have spent enough amount of time on arithmetic progression and some of arithmetic series now we'll talk about the geometric progression now we'll talk about the geometric progression so what's the geometric progression the metric progression is basically a sequence where the listing of the sequence is such that every succeeding term is obtained from the preceding by multiplying it with a a common ratio or multiplying it with a non-zero value which is called the common ratio so what is the basis of the sequence every subsequent term divided by the preceding is a fixed value which is a non-zero quantity okay and this non-zero quantity is called the common ratio what is this called common ratio okay now please note that this common ratio can never be zero this is important and also note that this implies that zero cannot be a term in any geometric progression okay please note that there cannot be any geometric progression you can say there cannot be any infinite geometric progression where zero is one of the terms right why because the term after zero if you divide it by zero that whole operation becomes undefined so zero cannot be a term of a infinite geometric progression okay even for that matter zero zero zero even if you write it people say sir then you know this is a geometric progression no here the common ratio r is undefined so it cannot be a geometric progression okay so if this is the you can say the characteristic of a geometric progression how do you find out the nth term of the geometric progression so let us use this so let's put k as a one so i'll get t2 by t1 as an r let's put k as a two i'll get t3 by t2 as an r if i put k as a three i get t4 by t3 as an r and let's say n minus one if i put i'll get tn by tn minus one as an r now this time multiply them all okay in case of an nth term of an arithmetic progression we added them here we'll be multiplying it so please when you know please don't that when you're multiplying it the numerator of this will get cancelled with the denominator of this and same will happen with all these terms which will leave you just with tn by t1 on the left hand side on the right hand side there will be r written or there will be r multiplied n minus one times with each other so that will become r to the power n minus one so from here you get tn is equal to t1 into r to the power n minus one and since you have assumed your t1 to be a the nth term of a geometric progression will be a times r to the power n minus one please note this down okay this is known to you because you have done this in school again a few points to be noted here a very important point just like in an arithmetic progression the sum of terms equidistant from the beginning and the end is a constant and that constant is the sum of the first and the last in the same way in the same way in a geometric progression please note down that the product of the terms the product of the terms equidistant equidistant from the beginning and the end is equal to the product of the first and the last term okay so let's say you're talking about a finite geometric sequence geometric progression okay let's say 1 2 4 8 16 okay and let's say I take one more term 32 the product of 1 and 32 should be same as the product of 2 and 16 should be same as product of 4 and 8 okay and please note that if at all there is a middle term it will be square of the middle term provided the middle term is there okay if it exists if it exists okay so for example if I write 1 2 4 8 16 32 84 sorry 64 so 1 into 64 is same as 2 into 32 is same as 4 into 16 is same as square of this okay or you can say sir 8 into 8 you can say 8 multiplied to itself so 8 square is this fine any questions any concerns apart from that also known down maybe I'll drag the screen to the right so please copy this down if you want to okay so please note down the following if you reciprocate the terms of an arithmetic sorry geometric progression please note that it is still in a geometric progression right this is still will be a geometric progression okay if you raise all the terms of a geometric progression by a power okay please note that this will also be a geometric progression product of corresponding terms corresponding terms of 2 gp of 2 gps will also will also result in a gp so if you have 2 gps let's say a a r a r square a r cube da da da da and there's another gp let's say b b r dash b r dash square b r dash cube da da da da even if you multiply the corresponding terms let's say a b then you have a b r r dash then you have a b r r dash square and so on and so forth this will also be in a geometric progression okay so 2 gps if you multiply the corresponding term it will also give you a it will also give you a geometric progression only okay so please note these down important properties might be helpful in solving few questions now let's take a problem to begin with can I take a question now done okay so here is a question for all of you a b c are positive integers forming an increasing gp okay so read the question carefully watch out for all the pieces of information provided to you a b c are positive integers positive integers forming an increasing gp increasing gp that's a keyword for us and b minus a is a perfect cube okay it's also given that b minus a is a perfect cube on top of that they've also provided that log of a to the basics plus log of b to the basics plus log of c to the basics is 6 when a plus b plus c is which of the following a plus b plus c is which of the following let me put the poll on for you okay so I have got three responses so far almost two minutes gone okay let's begin the countdown please vote in the next 15 seconds if you want to five four three one okay not many of you have responded anyways let's check this out so there is a tie between option c and d three three votes each and only one of you have said option b let's check which is right see the first information that is provided to us is that they are all positive integers okay so a b c are all positive integers we will keep this in our mind okay b minus a is a perfect cube this is one of the you can say very vague information which we cannot directly use but the next information that is given is a direct forward in a straightforward information where we can say if I have log a log b log c to the basics you can write it as log a b c to the basics as six which means a b c is actually six to the power six okay now a b c are in a b c are in geometric progression now please note that if you have to choose three terms in a geometric progression a b c always choose the middle one and the one which is right and left to it like this right at least use this when you have the information about their product so as you can see there is an information about the product of a b c so it's better to choose a b c like this now again this is experience so just to answer there just any question okay why is this experience is it illogical you know you can act no it is a logical move because when you multiply it your r's will get cancelled so when you write your a as b by r b as a b and c as b are you would realize that r and r here will get cancelled and getting b becomes very easy so from here you can get b as six to the power of two isn't it so at least you know that your middle term is 36 at least your b is a 36 okay fine so these terms are 36 let me write on top of it 36 36 r and 36 by r okay so these are your a b c now next let us use the last information which was a very vague piece of information b minus a is a perfect cube so b minus a is a perfect cube now perfect cube and b is more than a so it should be a positive perfect cube so it can't be zero also can it be one okay it could be one can it be eight it could be eight can it be 27 it could be 27 can it be 64 can it be 64 by the way of course not for that yours 36 minus r should be a negative quantity right so it can only be these three options correct one eight and six how karthik tell me six to the power six cube root is six to the power two no so can i say with respect to the with respect to the choices of perfect cube there are only three possible choices okay so this is the first choice second choice third choice okay now let us see which of the choices leads to the fact that i'm getting a b c all as a positive integer so if i take a one let's say choice number one if i take this as a one i will get 36 by r as 35 that means r will become 36 by 35 please note that if r is 36 by 35 your c term which is 36 r is 36 square by 35 which is not at all an integer forget about positive integer it is not an integer at all so this is ruled out so this is not possible okay next is eight if i take 36 minus 36 by r as an eight i will get 36 by r as 28 i believe so r is going to be 18 by 14 or you can say nine upon nine upon four yeah so my c which will be 36 r okay i hope my calculations are correct nine upon seven my bad nine upon seven so this is also not a positive integer so this choice is also gone for a toss right now next is 27 third case so if i take 36 minus 36 by r as a 27 it means 36 by r is a nine that means r is a four so c is equal to 36 r that is nothing but 36 into four which is going to be 144 yes this is an integer very much okay so this is accepted rest all of them will be rejected okay so your r is a four so after all we got b as 36 this is your b value c is 144 and since r is four your a is 36 by four which is nine so a is a nine so what's the question asking us question is asking us a plus b plus c a plus b plus c will become nine plus 36 plus 144 which is i think 45 plus 144 which is 189 okay which option is correct which option is correct option number d is correct is it fine so this is this is what is required to solve the question right so all the questions this chapter is easy but don't expect always straightforward questions from this chapter is it fine any questions any questions any concerns would you write the formula of log formula of log or the log property you want the log property which i use here okay the log property which i use here was log m to the base b plus log n to the base b is log m n to the base b okay and this could be extended to any any number of logs for example i have just taken two here but in the question they were three log a log b log c okay is this helpful uh well can we move on to the next question now can we move on to the next question if you're done with this next question is if a b c are real numbers such that such that thrice of a square plus b square plus c square plus 1 minus twice of a plus b plus c plus a b plus b c plus c a is equal to 0 then a b c are in option a arithmetic progression only option b geometric progression only option c both in arithmetic progression and geometric progression option d nota nota means none of the above and let me also conduct a poll for you here relaunch the poll once you're done please give me a response on the poll okay two minutes already gone only two people have responded yeah yeah please please take one moment no issues okay last 15 seconds post that we'll start the discussion five one i know why most of you have voted for b because you're you're you're thinking that pearly said is doing gp so you might have given a question on the gp okay okay let's uh let's see whether that was smart enough a move to make that decision based on the topic which has been discussed okay most of you as i told have gone for option b gp only okay let's check see here this term has been this expression that has been given to you has been given to you in a very you can say confusing manner okay let's try to give a format to this information okay i can see there are three a squares three b squares three c squares i can see there are three ones and i can see minus two a minus two b minus two c minus two a b minus two bc and minus two cx sitting in this expression okay so now what i'm planning to do i'm going to write it in a very interesting fashion i'm going to take one of the a squares i'm going to take a minus two a and i'm going to take a plus one and i'm going to create a perfect square like this okay and not only that i'm going to do the same with uh b square term as well c square term as well a square minus two a b plus b square term as well a square minus two c a plus c square term and b square minus two bc plus c square as you can see here i have utilized all the terms over here so as you can see three a squares have been used here one here one here one here correct three b squares will be used here same way this here this here and this here three c squares have been used here here here one three times here here here minus two a here minus two b here minus two c minus two a b minus two bc minus two c a okay so this term i've just written it in a slightly in a different way and this is basically nothing but sum of all squares as you can see these are all squares okay yes or no now a b c are all real numbers and the difference will also be real numbers so if you square the real numbers and there's some still comes out to be zero then basically it points out to the fact that each one of these should actually be zero so this should be zero this should be zero that means a should be one b should also be one c should also be one a should be b b should be c and a should be c in short in short a b c are all equal to one actually that means a b c a b c if you write them down they are one one one so what is this in ap or a gp or both uh what is this is it an ap or a gp or a both you tell me both isn't it it's ap also it's gp also okay and later on we'll learn that it will be in hp also okay so this is an ap with a common difference of zero this is also a gp with a common ratio of one okay so option number c is correct okay so googly question many of you who thought b will be the right answer because i'm dealing with geometric progression right now sorry negative one it was actually option c is it fine any questions any concerns this is an unconventional problem so please note that these these are the problems which if you do once you will be able to keep in your mind so there are two types of questions conventional problem non-conventional problem conventional problems you will be so able to solve it from the theory that you have learned non-conventional ones or unconventional ones once you solve one of those types you'll get an idea that oh this is this could be a possibility and you could apply it in the subsequent questions okay can i move on to the next concept do let me know once you're done with this great next we'll talk about a sum of n terms of a geometric series or a geometric progression okay so sum of n terms of a geometric progression which is a geometric series so let's talk about it so let me take a generic geometric series like this okay since we are only catering till n terms we will go to a r to the power n minus one now i'm going to use an interesting method to some of the series which was very close to the heart of one of the famous mathematicians of India shinivasa ramanujan okay he used to use this method quite a lot for summing many types of series in fact he used it to such an extent that he sometimes misused it and because of that he was pulled up by gh hardy who was supposed to be his mentor okay anyways so let's take this series and let's multiply this series with the common ratio r throughout that means each term multiply with an r after multiplying each term with an r just write it one shift it to the right that means multiply a with an r but write it like this okay right below this similarly multiply ar with an r and write below this ar square with an r right below this and so on this term if you multiply the term before this if you multiply with an r you'll end up getting this and this term when you multiply with an r you end up getting this right so what do you have done you have done nothing but you have multiplied every term of that particular geometric series with an r and just written one shifted to the right like this okay now why have I written it like this because now I'm going to take the difference of the two so subtract the two results okay so when you subtract the two results you get sn minus rsn which is actually sn times one minus r if you take sn common here if nothing is written you can put a zero by default here also you can put a zero by default now start subtracting these terms so a minus zero a start subtracting these terms this will go off for a toss start subtracting these terms this will also go for a toss start subtracting these terms this also will go for a toss go for a toss means cancelled okay and last term will be minus a r to the power n okay so from here you can make s in the subject of the formula this or this doesn't make a difference okay please note that I somehow come to know that in school some of the teachers are saying use this formula when modulus of r is less than 1 use this formula when modulus of r is greater than 1 there is no such distinction like that okay you can use any one of the formula for any of the scenario whether modulus of r is less than 1 greater than 1 right or for that matter equal to 1 also yes I'll tell you why it can be used for equal to 1 or how it can be used for equal to 1 okay so you can use any one of these formulas for any of the two scenarios okay so first make a note of this this is the sum of n terms of a geometric series now many people say sir you said you could use this formula even for r is equal to 1 how will you use that because it will give you a 0 by 0 right so if you want to use this formula for r equal to 1 right and actually I can use this formula for r I should be able to use this formula for r equal to 1 because when I was deriving it there was no such restriction I kept that r cannot be 1 right isn't it so I should be able to use this formula for r equal to 1 also by taking a limiting case please note that here I have to take a limiting case of any one of the two formulas let's say I use a second one r to the power n minus 1 by r minus 1 so here please recall that you had done a standard limit limit uh standard algebraic limit with me in the last class if you recall what is that limit x to the power n minus a to the power n by x minus a eight ending to eight ending extending to a okay so this this is basically same as the formula which we had seen in the last class what was it limit extending to a x to the power n minus a to the power n by x minus a is n a to the power n minus 1 is that here x is r a is 1 okay so nothing is going to change it is going to be n 1 to the power n minus 1 and of course this a is already sitting over there so it will become n a and I really should give you n a why because why because when your common ratio is 1 isn't your sum till n terms will be a plus a plus a n times okay so won't it become n a is it fine any questions any concerns okay so please note this down it didn't work saying that is not correct it worked but with a limiting case you have to take a limit value is it fine so this is how maths justifies it so that's why I say you have to take a limiting case in such scenario when r comes very close to 1 but not equal to 1 okay so limiting case is the answer to this scenario see I tell you something very interesting you know that the there's something called Euler line in in triangle Euler line is basically a line which connects the orthocenter the centroid and the circumcenter right and on the Euler line on the Euler line the centroid divides the orthocenter and the circumcenter in the ratio of 2 is to 1 okay sharduli but if the triangle becomes equilateral you already know that orthocenter circumcenter and the centroid will be lying at the same point so what happened to that ratio what happened to that ratio that 2 is to 1 they're all same point so where is that 2 is to 1 there so there comes the concept of limiting case so their distances is as good as a 0 but 0 by 0 is still 2 is to 1 because that is an indeterminate form is what we treat it as the this is how mass justifies these kind of scenario are you getting my point what I'm trying to say so you may be very adamant in saying that r has to be 1 only I don't want to hear anything else but in that case what will happen you will not be able to justify that sum of a plus a plus a plus a taken n times is undefined that is not correct right so there will be a you can say mismatch between the two results so this is how mass justifies that no both the results should give you the same answer but in one of the cases you have to take a limiting limit as r tends to 1 then only you'll get the results okay I know many of many of the you can say people don't agree with this kind of a approach but this is how mass justifies things you know in some of the extreme scenarios okay we say we say a asymptote is a tangent to a curve at infinity infinity never comes so it is never a tangent how can you say this tangent again it's a limiting case it's a limiting case okay anyways so let's talk about one more scenario over here when your r modulus is less than 1 okay in this situation in this situation you can actually find the sum of a geometric series even if the number of terms are infinite okay so when r is having a magnitude less than 1 that means when r is between 1 and minus 1 then then s n or you can say then s infinity can be obtained okay why because when your r magnitude is less than 1 it makes it makes the infinite geometric series infinite geometric series a convergent one a convergent one now what is the meaning of convergent convergent means even though you are summing it to infinitely many terms the sum will be a finite quantity that is called a convergent infinite series so we will not go into the depth of convergence topic because you are anyway is going to read it in your undergrad I think in your second semester or first semester you have a topic called test for convergence in your undergrad I remember in IIT I studied it in in my first semester only so there are a lot of tests for convergence limit test is there Rabi's test is there uh integral test is there p test is there all those tests are there anyways so when the series becomes convergent that means s infinity can be obtained so how do we obtain s infinity so again we'll take some limit concept here to get the answer but before that I would like to share a small theory with you if you have a quantity whose magnitude is less than one so if you have a quantity whose magnitude is less than one and you're raising that quantity to a very large power this result will ultimately be a zero okay so please remember this result this is going to be helpful in our limits chapter as well when we are doing it in more detail so when a term is having a magnitude less than one that means somewhere between minus one to one but not including minus one and one and if you raise that kind of a term to a very large power then the result will be zero you can try it out on your calculators also let's say half into half into half into half into half into half into half if you do infinitely many number of times you will eventually realize that your calculator will start throwing answers like 10 to the power minus 19 10 to the power minus 50 like that 10 to the power minus almost it will become a zero okay so exploiting this concept or rather using this concept we can say that if I want to find out the sum to infinite terms I have to take the limiting case of the result which we had derived as n tends to infinity so let me take any one of the result let's take this one so if n tends to infinity and r is having a magnitude less than one then this guy will become a zero giving us this which is as good as a by one minus r so please note that for a convergent infinite geometric series the sum till infinity is this note this though if your r is equal or greater than one or less than or less than equal to minus one please note that under that scenario your geometric series your infinite geometric series will not be a convergent one that means you will not be able to find out a finite answer for the sum till infinite terms so only when mod r is less than one it is possible to find out a finite answer even though the sum of the even though the number of terms in that series is infinitely many is this fine any questions any concerns we will be using this yes in one of the questions we'll be using it why not can I can I move on to questions now should I move on to questions okay so the first question is basically a simple one I am I'm sure you would have seen this question infinitely many times in school also find the sum of n terms of this series 5 plus 55 plus 555 and so on till n terms okay if you have already done it well and good okay you can relax but if you have not done it please try it out many people would have already done this in school and this comes in various shapes and sizes sometimes they'll say 0.5 0.55 0.55 or 7 plus 77 plus 77 like that I mean they can just change with change the numbers okay it could be five sometimes it could be seven sometimes it could be two sometimes depends upon the teacher please do not directly use the formula for the sum because it is not a geometric progression I hope you are able to figure that out it's not a geometric series I should say the right word is series if you're done you can just write it done on the chat box okay let's discuss it so for such kind of a question there's a typical approach we normally take five common okay let's say till n terms now what do we do we multiply and divide with a nine so instead of one eleven hundred and eleven we make we multiplied with nine okay and of course we compensate by dividing with a nine now what is the benefit of this this nine could be written as 10 minus one 99 could be written as 10 square minus one 999 could be written as 10 cube minus one and you can see there will be n terms like this so last term will be 10 to the power n minus one correct now take all the 10 terms together take the various powers of 10 together 10 to the power one 10 to the power two 10 to the power three all the way till 10 to the power n together and all these 11111 take it together so that will be minus n so now this term is a gp or these terms are in gp so we can find out the sum of these terms so that will be a r is 10 number of terms is n minus one by r minus one minus n so this will give us five by nine 10 into 10 to the power n minus one by nine minus n so this gives us five by 81 correct me if i'm wrong is this fine any questions now i'm sure you would have done this question in school as well any questions let's take another question by the way this question is a diagram please please watch out this diagram very carefully okay so there is a square there's a outer square abcd whose side length is one okay so this is a square so abcd is a unit square you could say abcd is a unit square unit square means a square whose side length is one okay side length is one okay now pqr are such points which divide this in this ratio please watch this ratio this ratio is alpha is to one minus alpha again alpha is to one minus alpha alpha is to one minus alpha alpha is to one minus alpha okay where alpha is some quantity which is between zero and one okay and this process is continued on and on my my figures might not look like a square but please consider them to be squares only so this is alpha one minus alpha alpha one minus alpha alpha one same ratio okay alpha one minus alpha and so on that means you can have this goes on and on and so on now my question to you is if ai represents the area of the ith square if ai represents the area of the ith square and summation of ai from i equal to 1 to infinity is 8 by 3 then find the values of alpha then find the values of so a1 is the outer most square abcd a2 is the next square apqrs a3 will be the another one you can name it bits by some alphabets maybe m n l r okay so that is a3 then there'll be a4 a5 a6 if it goes all the way inside inside inside infinity the sum of all the areas is coming out to be 8 by 3 what is the possible values of alpha question is clear to everybody any doubt with respect to the question do let me know oh i'm sorry uh you can you can change this to m l n you can call it as s if you want oh really but it would not have been as complicated as this maybe you know similar in line but not very complicated right yeah they they said midpoint okay see everybody please pay attention right now i've taken uh any two squares one within the other any two squares okay let's say let's say this is a square whose side length is small a n okay and this is a square whose side length is small a n plus one okay small n is the uh side length of the peak square you may call it as the uh nth square you can call it and this inside gray one you can call it as the n plus one at square so uh it is very obvious that as per the requirement this is alpha a n this is one minus alpha a similarly this is also one minus alpha a n okay now just focus on this triangle focus on this triangle i'm just making a bubbles around it okay now this is our right angle triangle whose three sides are alpha n one minus alpha n and a n plus one so can i say pythagoras theorem can be applied here so on this triangle i can say a n plus one square is alpha square a n square and one minus alpha square a n square right now what is the square of side length area so can i say this is area of the n plus one at square here also if i take a n square common this a n square is as good as capital a n that means area of the outermost square so what does it tell us it tells us that the ratio of the area of the subsequent i mean inside square to that of the outermost square is this term which is one minus two alpha plus two alpha square am i right am i right and this is a fixed value because alpha is fixed so what does it mean it means all the areas of the squares are in a geometric progression isn't it so this means that this implies that the area of the squares are in GP with common ratio as one minus two alpha plus two alpha square am i right any question related to it now if you have the sum till infinite terms given to you you will say it is the area of the first square divided by one minus r r is this in short area of the first square is one only and this term is nothing but two alpha minus two alpha square okay this ratio is given to us as eight by three how would how would the length of all the squares will be one what is that logic to just any i didn't get that let's say let's say alpha value is half let's say this is a midpoint okay so the next square will be having a side length which is exactly let's say i take a take a demonstration over a simple demonstration let's say i i make a square whose length is one okay and then i connect their midpoints this is one one one one if i collect their midpoints this is half that is 0.5 this is 0.5 so this is under root two into 0.5 no how is this one it becomes one by root two exactly no isn't it say yes no maybe this will become one by under root two okay and again if you connect their midpoints they will still become one by under root two of the same which is one by one by root two root two which is half anyways so once we have got this this is a simple quadratic that we will get this will be a simple quadratic that we will be getting in terms of alpha let's try to solve for alpha is this quadratic factorizable i i think so it is factorizable because 48 is 12 into 4 and i can see that happening over it okay so let's take 12 4 and take 4 alpha common this will be 4 alpha minus 3 minus 1 common 4 alpha minus 3 so 4 alpha minus 1 times 4 alpha minus 3 is 0 so that makes alpha value or 4 alpha value as a 1 or 4 alpha value as a 3 that means alpha could be 1 by 4 or alpha could be 3 by 4 okay both the situations are possible and you can see that why this both situations are possible because if alpha is one fourth then alpha and 1 minus alpha will be 1 is to 3 and alpha is 3 fourth then the same will be 3 is to 1 but that doesn't change the change the entire situation okay so both are possible so alpha values could be 1 by 4 or 3 by 4 is it fine any questions any concerns till this karthik or should i take it a little lower or higher okay i think after this step you should be able to solve it on your own just a quadratic to solve fine so with this we will take a small break on the other side of the break we'll talk about the means arithmetic means and geometric means okay so let's have a break right now i'm sure all of you must be hungry right now time is 6 o 9 we'll meet exactly at 6 24 p 6 24 p m okay little to the left top this is this is the whole screen after that whatever you have copied this is the whole screen okay enjoy your break see you on the other side of the break so i hope so next we're going to talk about means okay so i'll be starting with arithmetic mean first now the concept of mean is used in two ways uh you know depending upon how the concept has been framed in the question so one type of question is arithmetic mean of certain numbers so they say arithmetic mean off okay the word here is off so if you have let's say a1 a2 a3 okay till a n so arithmetic mean off these numbers will be some of these numbers divided by the total number of terms the same way as you used to find mean in statistics okay so arithmetic mean off i'm so sorry can you just see my screen yeah is my screen visible to everybody okay yeah so arithmetic mean off of n numbers is nothing but some of all the numbers divided by the total number of terms okay and there's another term that we use is insertion of arithmetic mean between two numbers insertion of arithmetic means between two numbers between two numbers this is more important to us because we'll be tested on this concept in this chapter so first of all what is the meaning of inserting arithmetic means between two numbers so let's say between any two numbers a and b okay if i insert let's say n arithmetic means a1 a2 a3 etc okay first of all what is the meaning of the statement that i'm inserting n arithmetic means between a and b it just means that all these numbers a small a capital a1 capital a2 capital a3 and so on so forth will be they are in arithmetic progression okay so all these numbers will be in arithmetic progression okay so if i've been provided with a b and i've been provided with the number of arithmetic means that i'm inserting between a and b i can actually find out what are those arithmetic means in terms of small a and small b let me begin with the process by asking you a simple question so let's start with inserting one arithmetic mean between a and b so let's say capital a1 is the arithmetic mean between small a and small b okay tell me what is the value of a1 in terms of a and b in short i'm asking you what is capital a1 if these three terms are in ap simple right you can say two a1 is a plus b okay everybody knows that if three terms are in ap then some of the terms equidistant from the beginning and the end is twice of the middle term okay so that is happening in this case also so a1 is equal to a plus b by 2 okay luckily a plus b by 2 is also the arithmetic mean of a and b also okay so it is the arithmetic mean between a and b also and it is the arithmetic mean of a and b also is it fine okay so let me extend the scenario if i have to insert two arithmetic means a1 and a2 between small a and small b okay tell me what is the value of capital a1 and capital a2 in terms of small a and small b find capital a1 and capital a2 in terms of small a and small b small a small b is given to you and karthik i'm asking you to insert two arithmetic means between a and b capital a1 and capital a2 so what should be the values of capital a1 and capital a2 okay all right so for this again there's no rocket science involved let's say the this four terms are in ap right so can i say capital a1 will be a plus d capital a2 will be a plus 2d and b will be a plus 3d so d will be b minus a by 3 okay so if you put these values over here and here your capital a1 will become a plus b minus a by 3 which is nothing but 2a plus b by 3 and capital a2 will be a plus 2b minus a by 3 which is nothing but a plus 2b by 3 is this fine now have you started noticing a pattern in which these am arithmetic means are basically written in terms of a and b if so then we'll take one more round so let's say i have a and b and i'm inserting three arithmetic means between a1 and a2 and a3 okay so what is your a1 what is your a2 and what is your a3 in terms of a and b please write this down everybody is it done see again same thing this is a plus d this is a plus 2d this is a plus 3d and d could be found out by the fact that b is your fourth term i'm sorry fifth term so it is a plus 4d so from here d is b minus a by 4 put it back over here put it back over here put it back over here so it becomes a1 as a plus b minus a by 4 which is nothing but 3a plus b by 4 similarly a2 would be a plus 2b minus a by 4 which is nothing but 2a plus 2b by 4 and a3 will be a plus 3 into b minus a by 4 which is a plus 3b by 4 okay so i think the pattern is very obvious from here the pattern is if you are inserting three arithmetic means okay your denominator is always one more than that as you can check four four four here you are having three three here you are having two okay and if you're inserting three arithmetic means the first arithmetic mean will be three into a one into b by four and slowly three will start diminishing and one will start increasing okay so this pattern if you are i know if you have recognized that you will be able to write it even for if let's say i'm inserting four arithmetic means between a1 a2 a3 and a4 okay so a1 here you can write it as 4a plus b by 5 a2 you can write 3a plus 2b by 5 a3 you can write it as 2a plus 3b by 5 a4 you can write it as a plus 4b by 5 okay so we'll do it in general so in general if you want to write it so if between two numbers small a and small b you are inserting you are inserting n arithmetic means so please note that a1 will be n a plus b by n plus 1 a2 will be n minus 1 a plus 2b by n plus 1 a3 will be n minus 2 a plus 3b by n plus 1 and let's say kth arithmetic mean will be n minus k plus 1 a plus kb by n plus 1 and this will go on all the way till a n a n will be a plus n b by n plus 1 okay please make a note of this okay so with this we'll take a question so please read out this question the question says between 1 and 31 there are m arithmetic means so that the ratio of the 7th and m minus 1th means is 5 is to 9 find the value of m so between 1 and 31 you are inserting m arithmetic means okay and a7 by a m minus 1 is 5 is to 9 you have to find the value of m so a7 will be m minus 7 plus 1 a plus 7 into b by m plus 1 correct if you simplify this it becomes okay similarly a m minus 1 will be m minus m minus 1 plus 1 into 1 into m minus 1 into 31 by m plus 1 okay so this will become 31 m minus 29 by m plus 1 so this two ratio is 5 is to 9 this two ratio is 5 is to 9 so let's find out m from here please solve for m and let me know your response correct they are spinning so basically this will be 9m plus 1899 this will be equal to 155 m minus 145 so 146 m is equal to 2044 so m is going to be 40 correct is it fine any question any problem with respect to the arithmetic means next question which is one of the very famous question asked in school exams also if a to the power n plus 1 b to the power n plus 1 by a to the power n plus b to the power n is the arithmetic mean between a and b where a is not equal to b find the value of n find the value of n i mean it's very easy to guess the answer also but i would like you to solve this via you know proper process okay so as per the question this is this should be equal to a plus b by 2 right okay let's say to solve the value of n from here so let's multiply let's cross multiply first of all this into this will be a to the power n plus 1 this into this will be b to the power n plus 1 and apart from that you'll get a to the power n b a b to the power n so this will go for a toss this will go for a toss so you'll have a to the power n plus 1 b to the power n plus 1 is equal to a to the power n b a b to the power n just take this term to the left side and this term to the right side so this will become a to the power n plus 1 minus a to the power n b this will become a b to the power n minus b to the power n plus 1 take a to the power in common okay so you'll get a to the power n a minus b on the left and b to the power n a minus b on the right just cancel off these two factors because a is not equal to b. So this will be a to the power n equal to b to the power n. This will be nothing but a by b to the power n equal to 1. So this means two things, either a by b is one, that is a is equal to b, but this is not possible because it has been provided in the question that a and b are not equal to each other. And the other possibility is that the power itself is zero. Okay. So this is the one which will help us to find the value of n. So n is zero in this case. Does it find any questions? Any questions, any concerns? So with this, we are now going to move towards geometric means. So again, with respect to geometric means, we have two types of terms being used. So let's say a1, a2, a3, till a and r positive terms. Okay. What do the geometric mean of these terms? So please note that geometric mean of these positive terms is defined as the product of all these terms raised to the power of one by m. Okay. But please note that this formula is to be used only when all these terms are positive. All these terms are positive. Okay. If it is negative, then you will have a problem finding out the GM. I will tell you why it is like that. See, if I have two terms a and b and both are positive. Okay. So both are positive terms. Then the geometric mean of these two terms is under root of productivity. But if both of them are negative terms, then it will be negative of under root of a b. Right. Now, why it is negative? Now, many people ask me, sir, why it should be negative? See, because let us say I take, for example, a as minus four, b as minus nine. Now, I want a term between minus four and minus nine, which would be in such a way that these three are in geometric progression. Okay. And that term is a minus six. And that is obtained by doing negative minus four into minus nine to the power of half, which is negative of six. Okay. That is why a negative sign comes. So depending upon the terms, you'll have to change your methodology. Now, why it is minus because mean is supposed to be between. It cannot be beyond it. It cannot be under root a b because under root a b will give you a six. Six is not between minus four and minus nine. It is beyond it. Okay. The word mean itself is something in coming in between. Now, I know the obvious question coming in your mind is what if a and b are of opposite signs? If a and b are of opposite signs, then please note that the geometric mean is not defined. Geometric mean is not defined. That is why when I wrote this generic formula, I gave a category mention of the fact that a one, a two, etc. delay and are all positive. Is it fine? The next type of question that is asked is insertion of insertion of geometric means between two quantities. So let me ask you this question if I am inserting one geometric mean between a and b, then what is the value? Let a and b are positive. Let me just categorically mention here that these are positive quantities. So if a and b are positive and I have to insert one geometric mean between them, what is the value of that geometric mean? In fact, I've already done with you a to the power half b to the power half also. Let me put two geometric means between a and b. Can you tell me what should be the value of g1 and g2? See, again, what is the process involved? Let's try to understand this. The process involved is that these are in geometric progression. So the same way as we dealt with our arithmetic means, we have to deal here also. So I have to say g1 is a r, g2 is a r square and you can say b is your a r cube. So r is b by a to the power of one-third. So if you put it in the expression of g1 and g2, you will realize that g1 will become a into b by a to the power of one-third, which is a to the power two-third, b to the power one-third and g2 is going to be a into b by a to the power of two-third, which is nothing but a to the power one-third, b to the power two-third. Please note the pattern here also, especially with respect to the powers that you are raising on a and b. It is very similar to the coefficients of a and b, which we had seen in our arithmetic progression. Does it find any questions? So let me ask you to do one simple exercise here. Insert three geometric means between a and b. Just reply this question based on the pattern only. You don't have to solve it separately. I hope you have all written it down on your respective notebooks. g1 will become, tell me a to the power. What will be the power of a? Just tell me the powers. I just want you to tell me the powers. No need to write all the expression. So for g1, what should be the power of a? Power of three by four. b will be having a power of one by four. So slowly there will be a transition of the powers like this. Okay, so let us generalize this. So in general, if you have a and b and you are inserting n geometric means between them, g1 will be a to the power n by n plus one, b to b to the power one by n plus one, g2 will be a to the power n minus one by n plus one, b to the power two by n plus one, g3 will be a to the power n minus two by n plus one, b to the power three by n plus one, gk will be a to the power n minus k plus one by n plus one, b will be k by n plus one, and gn will be a to the power one by n plus one, b to the power n by n plus 1. Is it fine any questions? I hope you have all noted it down. Question is if a to the power n plus 1, b to the power n plus 1 by a to the power n plus b to the power n is the geometric mean, the same question but I am now changing the arithmetic mean to a geometric mean. Is the geometric mean of a and b where a is not equal to b, find the value of n. Find the value, any response? So you are claiming this to be your arithmetic mean of a and b. So cross multiply. So cross multiplication will give you an a to the power n plus half, b to the power half, then a to the power half, b to the power n plus half. Take this term to this side, take this term to the other side. So you will have a to the power n plus 1 minus a to the power n plus half b to the power half. This will be a to the power half, b to the power n plus half minus b to the power n plus 1. So this will be a to the power n plus half a common, it will be a to the power half minus b to the power half. Here also you can take b to the power n plus half common, it will be a to the power half minus b to the power half. Okay, now these two terms cancel them off because they will not be equal to each other. Okay, so because a and b are not equal to each other a minus a to the power half minus b to the power half will not be zero. So they're non zero terms, so they can be cancelled off. So from here, I can say that this is equal to one. So again, two possibilities, a could be equal to b, but this is not possible. Or the other possibility is n plus half is zero, which means n is equal to minus half. Is it fine? Any questions? Yeah, let's take this question. Easy. So let's discuss this. So a is the arithmetic mean of b and c. So a is b plus c by two or b plus c is two a. Okay. And g1 and g2 are the geometric means between the same two quantities b and c. Sorry, it was b and c not a and b and c. b and c. So g1 will be what? g1 will be b to the power of two by three c to the power of one by three. That means g1 cube will be b square c. g2 will be b to the power one by three c to the power two by three. So g2 will be g2 cube will be b into c squared. Okay, now as per the question, we had to prove that g1 cube plus g2 cube is equal to two a b c. So let's figure it out. What is it? So g1 cube is b square c. g2 cube is b c square. Take b c common. It is b plus c. And as per this expression, b plus c is two a. So you can use it over here. So b c into two a that will become two a b c, which is your right hand side and stroke. Is it fine? Any questions? Prove that if a1, a2, a3 till a n are the or you can say if if a1, a2, a3 till a n are the arithmetic means between a and b between small a and small b, g1, g2, g3 till g n are the geometric means between a and b. Then prove that sum of a1, a2 till a n is n times a plus b by 2. That's number one proof. Number two proof is product of g1, g2, g3 till g n is under root a b to the power of n. That means it is n times the a m of a and b. And this is g m of a and b to the power of n. Let me know once you're done with the first part. Okay, so let's discuss this out first one. See what is a1? a1 is n a, sorry, n a plus b by n plus 1, correct. a2 is what n minus 1 a plus 2 b by n plus 1. And this continues till a plus n b by n plus 1. So when you add them all, when you add them all, what do you get? Take a terms together and you will have a by n plus 1. You will have 1 plus 2 all the way till n. Okay, so just take this part, a's, a's together. Okay, I'm just really drawing it. So just this part a divided by this n plus 1. Similarly, take the b part, it will be 1 plus 2 till n. So this will be nothing but n into n plus 1 by 2. This will also be n into n plus 1 by 2. So n plus 1, n plus 1 go on, n plus 1, n plus 1 go on. So this will be n by 2 a plus b, which is nothing but n times a plus b by 2. Okay. In the similar way, I think the same thing will come as a power on a and b also. So when you take the product of all the geometric means, so it will become a to the power n by 2, b to the power n by 2. So what will happen with the coefficient? The same will happen with the powers here. So it is nothing but a to the power half, b to the power half to the power of n. So that's nothing but root a, b to the power of n. Is it fine? Any questions? Now, these two questions, which I gave you, they're actually properties of a, m and gm. So please, make a good note of this because they are sometimes asked as direct questions in comparative exams. Next, a, m, gm inequality. See, if a1, a2, a3, they're all positive terms. Okay. So these are positive real numbers. So let me not write if I just say a1, a2, a3, till an are positive real numbers. Please note that the arithmetic mean of these numbers, which is a1 plus a2 plus a3 divided by till an divided by n, this would always be greater than equal to their geometric mean. But one thing to be noted here is that this is only applicable when these terms are all positive real numbers. So only for positive quantities, this is applicable. So a, m is always greater than equal to gm. Okay. And equality holds, equality holds when these terms are equal. Equality holds when these terms are equal. Okay. Now many people say, sir, can you prove this? I'll prove this in general when I'm doing this chapter once again with you in more detail. But as of now, I'll prove it for two terms. Okay. So let's take a1 and a2 to be just two positive real numbers. Now, we all know this basic inequality that everybody would agree with that this term square will always be greater than equal to zero. Now, you would be wondering why did I choose this term? Why did I choose this inequality to prove this? Because if you square this, you will end up getting a1 and a2. And you will also get under root a1, a2, which is basically what is required for proving or for finding your arithmetic mean and geometric mean. So this means a1 plus a2 is greater than equal to under root a1, a2. So a1 plus a2 by 2 will be greater than equal to under root a1, a2, which means arithmetic mean would be greater than equal to the geometric mean. And you can see for yourself that if a1 and a2 are equal, what will happen? a1 plus a2 by 2 will become a1 only or a2 only, whatever you want to call it. And a1 into a2 under root will become a1 square under root, which will be a1 itself or a2 itself, whatever you want to call it. So they both will become equal to each other. They both will become equal to each other in case of this. Is it fine? Any questions? Any concerns? That's what I was saying that people will ask me why did I write it like this? Because if you expand it, you end up getting a1 plus a2 also in that term and you get under root of a1, a2 also in that term. So that was basically obtained from a reverse engineering point that I wanted a1 plus a2 and I wanted under root a1, a2 also. So that could have happened only when I would have taken under root of a1 and under root of a2 into my consideration. So reverse engineering, I got that. So I think this is all we need to know about sequence series and progressions for your semester exams. In fact, the chapter is not over yet. We would have to do more things after your first semester exam.