 sequences and series sequences and series okay I'm sure most of you would be doing this chapter in school or would have done this chapter in school if you've done this chapter in school how did you find it easy moderate difficult easy okay easy Shashod not good okay by the way many books will also call this chapter as progressions and summations progressions and series okay series or summations whatever you call it fine so I'm getting a mixed response people found it easy some people found it not that good okay anyways before I start this chapter since you have done this in school I would like you to answer me a simple question what do you think is the difference between sequence a progression and the series I would love to hear from you since you've done this chapter in school what according to you is the difference between these three terms are they the same right or they're different if they're different how they are different anybody if you want to speak out also what is the difference between sequence progressions and series seek progression there is a common difference okay sequence is like an ape is like AP okay okay so if I ask you what is a human being human being is like akhil is that how you will answer whatever sequence is a pattern sequence is a bunch of numbers following a rule okay predefined rule progression is change sequence is any random collection of terms okay so any random correctional terms will become a sequence okay series is the sum of a sequence okay now I'm getting mixed response from people okay some of your arguments are correct some of your arguments are not okay now let me take this over okay thank you for your opinion love to see your responses see sequence is a listing of is a listing of numbers it could be alphabets or letters it could be phrases okay where there is a pattern or a rule a well-defined pattern or a rule okay it cannot be random okay there has to be a rule or a pattern then only we'll call a sequence I don't know how many of you have written NTSC in NTSC there was a segment called mental ability test in mental ability test the very first chapter is sequence okay so if we have to find out the next letter in the sequence or next number in the sequence or for that matter next word on the sequence right so sequence is a listing where there is a pattern or a rule hidden that's how we can find out a subsequent term of a sequence okay progression is a sequence so progression is what it's a type of a sequence where there is an explicit formula there is an explicit formula formula to find the value of the next term or to find oh sorry I'm writing it twice to find the value of a term from its position now what is the difference between these two let me give an example for this okay series I will come to in some time I'll just give the example and cite the difference between see let's say I give you a listing like this two three five seven nine eleven thirteen seventeen nineteen twenty three okay I'm sure you would have recognized this what is this this is a sequence of this sequence of prime numbers right absolutely I think but we'll not call it as a progression because there is no formula for finding the value of a prime number given its position is known for example if I say give me 8764 prime number will you be able to find it out does that does there any formula exists which will tell you that okay if you put the position of that prime number throughout the value of that prime number is there any formula where I'll put let's say five and I should get oh sorry for for writing a nine sorry okay let's say if I put a five I'll get a 11 is there any formula like that or if I put let's say 10 I should get something like 23 or let's say the next prime number is 29 okay will I get like that there is no such formula so we will call this as a sequence of prime numbers not as a progression of prime numbers so this is an example of a sequence of prime numbers okay so there is no explicit formula by which we can know the value of a prime number given its position is known to us but what is a progression progression on the other hand is those sequences for example if I write something like this 147 1013 16 da da da da now in this also there is a pattern but I know that there is a formula for the nth term which in this case is 3 and minus 1 or sorry 3 and minus 2 okay so if I put n as 1 I'll get the first term if I put n as 2 I will get the second term if I put n as 3 I'll get the third term if I put n as 4 I'll get the fourth term so there exists a formula by which I can generate the value of a term if I know its position so such sequences will be called as a progression right now somebody who said there is a common difference see those are all very limited examples you are giving progression is a much broader term you could have a geometric progression you could have a harmonic progression you could have an arithmetic geometric progression right so it's not only about common difference only the main criteria for you to call a sequence as a progression is that there should be a formulation for finding out the value of a term given its position is not in sequence such formulations may not be present okay so by this logic you would have understood that sequence is a bigger set and within that bigger set there is a set of there is a set of progression okay so let's say this red circle represents sequences then this yellow circle will represent progressions so progressions are types of sequences okay all sequences formulation is not possible but yes there is a pattern hidden in it in progressions there is a formulation hidden okay now just a small question for you one one two three five eight thirteen twenty one data data okay have you heard of this listing before see the wrong use of words Fibonacci series it's not a series for sure it is a Fibonacci sequence correct is it eligible to be called a progression yes because there is a formulation made for Fibonacci sequence okay and that is called the Bennett's formula I would request everybody to search it out Wikipedia YouTube whatever sources you can Google it out there is a formula no this is not a formula this is a relation this is a pattern Anurag Anurag you are Anurag is mentioning this Anurag this is a pattern by which it is built up but there is a formula for finding the NET term figure it out okay it has some got irrational terms 1 plus root 5 by 2 etc will be there that formula is called the Bennett's formula so this is actually qualifying to be called as a Fibonacci progression also right but mostly people who do not know this formula they play a safe hand they say Fibonacci sequence but Anurag not series mind you okay so what is series series is when you add terms of a sequence so series is a summation so when you add the terms of a sequence add terms of a sequence that becomes a series okay so when sequence terms are added up you can say subtracted also see subtraction is a special form of adding adding of negative terms right so sequence when they're added that will be series so don't use don't interchangeably use these words please do not interchangeably use this word who is spamming front of no permutation of ways of adding 1 and 2 to get I didn't get that no permutation or number of permutation I didn't understand that at all sorry for that my daughter yes I didn't understand what you meant to say by that line okay anyways is the difference clear now see guys whenever you're talking to you know people who are have a bit of knowledge in maths please use proper words right don't say Fibonacci series for it I know many people are using wrong words right this is not a series unless until you add them it is not a series okay not all sequences are progressions however you can use sequences for progressions also that's why this chapter name is sequences and series because progressions are types of sequences only but remember all sequences need not be formalized they may not have a formula in place okay anyways so with this knowledge with this knowledge we will understand what are we going to study in this chapter so let me give you an overview okay in this chapter we are going to first talk about three different types of progressions okay we are going to talk about three different types of progressions what are they arithmetic progression which is something you have already done in your class 10th and have you have you have all you know done exceedingly well in this chapter because this is supposed to be a very easy chapter you will be introduced to geometric progression geometric progression okay and there will be one more type of progression called the harmonic progression which is not there in your school syllabus so if your school teacher would have taught you this chapter she would not have touched upon harmonic progression because this is not a part of your cbsc curriculum second thing that we are going to learn in this chapter is the concept of means okay so three types of means will be there arithmetic mean geometric mean harmonic mean you have already heard of these terms we have done a few of the concepts prior to this by use of the relationship between a m g m and h m okay and very importantly the relationship between a m g m and h m the inequality between or people call it as a m g m inequality let me write it like that only a m g m a m g m h m inequality okay sir will you be teaching harmonic progression yes why not that's why I'm listing it out it is nothing it's very small it's just 15 minutes concept okay third thing that we are going to talk about is your series series so in series what are we going to learn we are going to primarily learn methods of summation methods of summation okay under this we are going to learn three concepts sigma method of summation method of difference and under method of difference we are going to specialize in one of the methods called the vn method vn method of summation and second thing we are going to learn is your a g s a g s a stands for arithmetic or g stands for geometric s stands for series okay so this is our entire we can say the index of the content for this chapter it should not take us more than two classes okay two classes or slightly more than two classes so we'll be able to finish it off meanwhile I would also like to discuss with you so let's start with arithmetic progression okay now this concept is quite well known to you all so I'll be fast here so that we save a lot of time in fact we'll spend time doing some questions so what is an arithmetic progression it's a sequence where a subsequent term differs from the preceding by a common difference okay so the subsequent term differs from the preceding by a common difference that means your n plus one-th term differs from your nth term or for that matter you can say oh it's fine n plus one is also fine this difference is always fixed so any kind of a sequence which meets this criteria would be eligible to call as a this is the pattern that that particular sequence is following right so this is the pattern now why am I calling it as a progression is there a formula possible for nth term or n plus one-th term let's see so there are two ways you can figure out the formula one way is if you see this is your first term this is your second term this is your third term this is your fourth term are you able to relate the position with the number of d's that you have here for example if this is first term there's no d or you can say zero d there is second term there's one d third term 2d fourth term third 3d right so by looking at it you can say that for your nth term there will be a plus n minus one d and there you go this is your formulation or this is your formula for your nth term of this particular sequence and hence this sequence qualifies to be called as a progression now this is by observation let's say somebody says sir i don't want to observe and give the formula can we use this pattern to come with a formula yes that is very simple start putting some values of n over here let's say i put n as 1 if i put n as 1 i get t2 minus t1 as d if i put n as 2 what do i get t3 minus t2 as d if i put 3 what do i get t4 minus t3 as d if i keep on going and put finally your n value as n minus 1 i will get tn minus tn minus 1 as d let's add them all let's add them all if you add them on the left hand side you would realize that there would be a cancellation of these terms happening okay as a result you will be left with tn and t1 and you would write d d d d d d d how many times n minus 1 times remember you started with 1 and you went till n minus 1 right so your tn will be t1 which you already know is a n minus 1d so this is another way to get to the formula for the nth term of an arithmetic correlation right this is already well known to you okay let's directly jump to questions because i think most of you are already aware of these so time to take a small quick question on this let's take a question okay i'll start with this question consider two arithmetic progressions s1 and s2 s1 has terms like 2 7 12 17 da da da da till 500 terms s2 has 1 8 15 22 till 300 terms there are two parts to this question one is find the number of common terms that's the first part second part is what is the last common term what is the last common term i would request everybody to use their last year experience and of course this year also you have done it i'm sure you will get this right so i want two answers what is the number of common terms and what is the last term so in your chat box your answer should look like two numbers separated by comma don't give me a single answer okay i want the last i want the number of common terms and i want the last term okay let's have around two and a half three minutes i think that is good enough for time come on guys first question let's do it give me two answers okay good okay aditya okay let's discuss this guys uh i think it has been quite a while now so uh let's write down these terms 1 7 12 17 22 27 da da da da okay now i would like to check whether the 500th term of the first series is less or whether the 300th term of the second series uh second sequence is this sorry for using the word series second sequence is less okay so let's figure out what is the 500th term of the first sequence so here the sequence is basically following a arithmetic progression the first term is two common difference is five so 500 minus one into five i think this will give you 202500 minus three right it's 2497 okay so the 500th term is 2497 and for the second sequence this is your s1 for the second sequence let's check 1 8 15 22 i think it's a common difference of seven so next term will be 29 da da da da 300th term what is the 300th term in this particular sequence so n uh sorry a plus n minus one n minus one is 300 minus one into d d is seven so this is going to be 2100 minus six which is 2094 okay so guys uh we have now first figure out that okay this guy is the smaller of the two okay this guy is the lesser of the two so nowhere no way i can exceed this number okay anyways now if i start finding out which is the first common term the first common term you can see is 22 okay so if i start making us a sequence of the common terms the first common term will be 22 can somebody tell me when will be the next common term occurring or what should be the you know uh difference between the next term and this term any idea 35 absolutely so the next term will occur at an LCM of five and seven okay see it's like saying that there are two bells in your school let's say one bell is ringing after every five minutes one bell is ringing after every seven minutes if let's say both of them ring now when will the ring next they will ring next after a gap off which is a LCM of five and seven LCM of five and seven is 35 we all know that so the next term will be 57 okay again next term will be 92 like that but remember whatever terms i get this term must be less than or equal to 2094 right so let's figure it out so tn should be less than equal to 2094 let's see where all we can go so that n will give me the number of common terms so 22 plus n minus 1 into 35 is less than equal to 2094 so n minus 1 into 35 is less than 2072 so n minus 1 by 35 will give me or you can say this will give me a number less than 2072 by 35 this is very close to 60 i think it's 59 point something so n minus 1 is less than 59 point let's say how much is this 59 and 0.2 okay so n is less than 60.2 so your n being a natural number can only be 60 so absolutely correct janta who said there will be 60 common terms and let's find out what will be that last term common so the last term common will be 22 plus n minus 1 into 35 okay how much is this 35 into 6 gives me 1 210 210 means 2100 minus minus 13 i believe so this is 2087 so this is your last common term aditya adrija i don't know how are you getting 2062 both of you counted it wrong okay is it fine any questions silly mistake adrijals okay never mind let's move on to the next concept immediately i think i don't i don't want to waste time some of n terms some of n terms of an ap of an ap so before we start this process i would like to tell you a small story a small inspiring story okay there was a small chap in germany i think a six-year-old boy in germany who was given a punishment by his school teacher to add up all natural numbers from 1 to 100 okay i don't know whether you have heard of this story so six-year-old seven-year-old probably not more than that and imagine a six-year-old guy given a you know a task of adding numbers from 1 to 100 how painful it will it be for this the student right but this guy was very smart what he did was he wrote the same set of numbers in a reverse order something like this okay and then he added both the series so s plus s is to s and when he started adding these numbers he realized that he's always getting a 101 see so when you're adding these they are always giving you 101 till the last number also 101 okay so he realized that okay he is writing 101 100 times so this into 100 so his s was nothing but 101 into 100 by 2 which is nothing but 101 into 50 and he basically found out the answer in quick time right so his punishment that was given to him basically was no punishment for him he actually could solve it in a you know minimal efforts this guy later on went to become one of the greatest physicists and mathematicians of all times Carl Frederick goss Carl Frederick goss i'm sure you would have heard of his name goss goss has given wonderful theorems in number theory he has done a lot of work in the field of electrostatics i think in class 12th the rest i will teach you about the goss theorem okay so this guy later on meant to become see six-year-old guy doing this right you can well imagine how far he can think of so the same concept that he used will be using also to find the sum of n terms of an ap so why i told you this story is because i'll be using a similar concept to find out the sum of n terms of an ap so let's say if you have a arithmetic series given to you like this now see i'm using a word arithmetic series because i'm adding terms of an arithmetic sequence or arithmetic progression okay so let's say i'm adding terms of this arithmetic progression from a to a plus n minus 1d so what i'm going to do is i'm going to write the same in a reverse fashion okay the second last term will be a i think is the last term actually so i could just write a name okay now start adding the two series in this fashion so add sns which is 2s add these two you'll get 2a plus n minus 1d okay add these two you will again get a 2a plus n minus 1d you can check it out add these two you'll again get a 2a plus n minus 1d and you will see that constantly your sum is fixed at 2a plus n minus 1d okay so how many 2a plus n minus 1d's have you written on the right side n number of 2a plus n minus 1d's right correct so this makes our life very easy so the sum till n terms will be nothing but n by 2 2a plus n minus 1d and this is the formula that you have actually learned in your class 10th also to find this okay now i have us another method to do this we'll try to figure out the same formula by another method okay so first make a note of this or that method i will talk about later little later on don't worry okay as of now let's keep this as our method that method we'll talk about when i'm doing series else there will be a repetition of concepts now there is there are a lot of takeaways from this formula it's not like you know you know this formula and we know everything about it no there are a lot of takeaways from this formula and those are very important takeaways the first takeaway i don't know how many of you observed is when you add the first in the last term it is the same as the sum of second and the second last term it is the same as the sum of third and the third last term and so on what does it mean it means that in an arithmetic progression the sum of terms equidistant from the beginning and the end guys the sum of terms equidistant from the beginning and the end that means kth term from the beginning and kth term from the end will add up to give you the same value as the first and the last term right k could be any number from 1 to n right so this is a very very integral property of an arithmetic progression every arithmetic progression will satisfy this some of the terms equidistant from beginning and the end will be same as the sum of the first and the last term okay and if at all there is a middle term it will be twice of the middle term okay i'll give you an example for this let's say i take a simple one one three five seven nine okay let me take more eleven thirteen okay one plus thirteen will be same as three plus eleven will be same as nine plus five will be same as twice of seven or seven plus seven you can say okay double of middle term is what adding the middle term to itself okay this is a characteristic satisfied by all arithmetic progressions okay hence we say that if we have three terms in arithmetic progression then a plus c will be twice of the middle term okay so a plus c will become 2b this is something that you have been already been using in tenth also quite a lot okay second important takeaway point is if at all this is the case i can always write this term as or sum of n terms as not only this we can also write it like sum of the first and the last term into n by 2 because this is actually made up of a and a plus n minus 1d so this is your t1 and this is your tn you can also write it as t2 plus tn minus 1 you could also write it as t3 plus tn minus 2 correct you can also write it as t kth term from the beginning and kth term from the end is this fine any questions regarding this i'm sure you would have already known this fact but the second fact is very important and some questions have been made on this any series any sequence any sequence whose sum till n terms is a quadratic in n of this nature will always represent will always represent a arithmetic progression please note this down okay but if your series is like this it will not be an arithmetic progression let me write apm guys this is very critical why this critical i'll tell you see we had derived this formula a little while ago right okay now many people when i ask them what kind of an expression is this in n they will say so it is a quadratic in n agreed absolutely no no problem with that particular statement but this is a quadratic where there is no constant term involved so many times this observation is not done by students so a question may come in the exam there is a series which satisfies this kind of a relation for the sum of that particular sequence will that sequence qualify to be called an arithmetic progression for this the answer is yes for this the answer is no unless until your c becomes 0 but if you let's say abc are non-zero quantities then this will not be qualified to be called arithmetic progression right in an arithmetic progression the sum of the terms will always be a quadratic in n without a constant term this is very important constant term should not be there are you getting my point i'll give you an example are you space is not there okay i'll give you head so let's say i give you that there is a sequence by the way this is a notation for sequence i don't know people who are prepared for olympiads and all this is a notation that we use for sequence okay so this sequence sum this sequence sum is given by let's say n square plus n plus 1 my question is is this sequence and arithmetic progression i know people will say sir let's find out the terms but that will take time isn't it for example let's say if i want to find out the first term first term is what s1 only so that will be three okay second term will be what second term will be s2 minus s1 yes or no s2 minus s1 will give you the second term s2 is what i think four plus two six six plus one seven seven minus three which is going to be four t3 will be what t3 will be s3 minus s2 okay s3 will be what 9 plus 3 12 12 plus 1 13 13 minus 2 which was 7 which is 6 okay here itself you can see that now the difference here was 1 now the difference has become a 2 right so by doing all these things you realize that boss this is this is not the sum of an AP okay but the moment i say moment i say there is a sequence whose sum of n terms is n square plus n is this a arithmetic progression immediately you can say yes it is has to be okay you can check it out t1 will be whatever you can say let's say t dash i'm writing t dash 1 will be what 2 correct t dash 2 will be what put a 2 here and subtract a 2 so 2 here will be 4 plus 2 so it'll be 4 okay t dash 3 will be what put a 3 here and subtract 6 so you'll get 9 plus 3 12 12 and minus 6 12 minus 6 is 6 yes you can directly see that it is coming out to be an AP it's coming out to be an AP so in nutshell what did i wanted to convey i wanted to convey here is the fact that if the sum of a given sequence or you can say the sum of a series is given to you as a quadratic without a constant you can immediately go for the fact that it would represent an AP but if the sum of a sequence or a series is given to you whose sum is containing a quadratic with a non-zero constant term immediately you can reject that it is not an AP don't waste a single minute also on that is this fine any question any concerns do highlight no questions okay so if you don't have a question let's solve questions okay here you go uh as my first question the sum of n terms of two APs are in this ratio what is the ratio of their 24th term and rock constant cannot be in terms of a changing thing n is a changing n is the number of terms how can a constant be n constant means a fixed number find the ratio of the 24th terms of two APs the ratio of whose sum till n terms is given to you as 3n minus 13 by 5n is plus 21 come on guys you've done these kind of problems in class 10th remember your childhood days when you were very small in class 10th yeah 11 12 is like totally different and you know what just one year from now you'll be filling the J form one year is already gone you're only come almost completed class 11 time flies and the sad part is we have not even got to see each other physically I hope you get a chance to do so in this year you can see me right I'm dancing on the camera only but I have not got to see many of you done very good Prateek very good aditya good good good anybody else anybody else okay and right good let's try this out see let's say the sum till n terms of the first series is sn and the sum of n terms of the second series is s dash in okay there are two different APs and let's say a is the first term and d is the common difference of the first sequence and a dash is the first term and d dash is the common difference of the second very good aditya n by 2 n by 2 will go off okay and this is given to us as 3n minus 13 upon 5n plus 12 sorry 21 not okay now what I'm going to do is I need the ratio of 24th term of both the sequences right so what is t24 of an AP a plus 23d and this guy will be a dash plus 23d dash isn't it okay now how do I get this without knowing my a and d so what I'm going to do is in this in this particular expression that I have I'm going to divide this by 2 this also by 2 note that I'm dividing by 2 both numerator and denominator so it is not going to affect anything so it's going to become something like this correct me if I'm wrong okay and this is still 3n minus 13 by 5n plus 21 now if you try to compare this chap with this chap you realize that if I replace my n minus 1 by 2 with 23 our work will be done isn't it so this ratio I could easily figure out isn't it so what I'm going to do I'm going to substitute n minus 2 by 23 in short I want to substitute n with 47 correct so a plus 23d a dash plus 23d dash here also I will substitute by 47 so 3 into 47 minus 13 upon 5 into 47 plus 21 okay I think it's an ugly value uh how much is this uh 141 141 minus 13 128 this is going to be uh 5 225 235 plus 21 256 oh yes answer is 1 by 2 brilliant adrija very good so adrija aditya prathik prathik is back into action prathik are you out of your hibernation you are maintaining a very low profile from the last few classes okay good let's move on to the next yeah the question is the sum of the first find the sum of the first 24 terms of an ap a1 a2 a3 etc if it is known that a1 plus a5 plus a10 plus a15 plus a20 plus a24 is 225 so you have been given a very very you can say a specific relationship between the terms sum and we know it's an ap we have to find the sum of 24 terms could you go back for one second okay after this follow and rock guide three very good excellent good pranav very good you guys are right oh you got it pranav good oh sorry and rock very good prathik advik okay now many of you would have actually substituted a5 as a1 plus 4d a10 as a1 plus 90 etc and i'm sure you would have got the answer by that but i would like to solve this problem in a slightly different way something which is related to the concept of the fact that the terms equidistant from the beginning and the end would be same as the sum of the first and the last term right so if you have let's say a1 that is the first term and the last term okay this sum will be same as the sum of a5 and a20 because a5 is the fifth term from the beginning and a20 is the fifth term from the end so these two will add up to give you the same thing similarly a10 and a15 are 10th term from the beginning and 10th term from the end respectively so they will also add up to give you the same thing right so the good part is when they give this to you as 225 they have actually given you let's say i call this as sum as x so they've actually given you 3x is equal to 225 that means x is equal to 75 right if you want the sum till 24 terms you can directly use the fact that n by 2 sum of the first and the last term which is 75 which is nothing but 12 into 75 900 is that clear if you can tell me an easier way to solve this i would be happy to accept it is it fine any questions any concerns please do ask okay so very strategically they had given us such terms which were like equidistant from the beginning and the end so first and the last fifth and the fifth last 10th and the 10th last okay find the middle term and multiply by n do you think Pranav that is a simpler way to solve it as compared to this i said if you have a simpler way to let me know there are several ways to solve it by the way okay let's take another one let us take another one another one another one okay if n the set of national numbers is partitioned into groups okay so they are made sets out of it s1 contains only one s2 contains two and three s3 contains four five six da da da da da find the sum of the numbers in s50 so if i continue making this partitions i will reach s50 what will be the sum of elements of s50 is what they're asking us try this out take two and a half minutes i don't know why your school is conducting the exams so early they should calibrate it with this year board right i don't know they have their own management and policies our arms and they could have conducted march end you know i know that uh charen normally all the schools have this that there is finish the syllabus by November and after that revision will happen and pre boards second pre board practicals project submissions lab records all those things will start so you all have to submit project reports i believe right in all the subjects including maths start working on it lab record also keep it up to date as what will happen last moment you'll start you know copying copy pasting from your friends and start bunking school and classes for all those things don't do that i know charen so your exams will be offline gaitri gaitri you come from this school dps i cheer okay panna good are you sure it is correct check money okay let's try to solve this problem see here if you see keep observing these things so they are something which is very very interesting if you see the last number of these sequence of these sets okay you would realize that this one is one of course this three is actually one plus two correct aditya correct that's correct this six is actually one plus two plus three correct so this gives you a lot of insight as to if i had gone till s 50 and i would have written all the elements in it my last element would actually have been one plus two till 50 correct i mean there are other elements also which i will eventually find out or you know i'll work around it don't worry but this guy would be the sum of all the numbers from 1 to 50 correct and we know the sum of numbers from 1 to 50 is 50 by 2 first term plus the last term which is nothing but 25 into 51 how much is that i think it's 12 75 correct so this guy at least you know is going to be 12 75 so the next guy would have been 12 74 12 73 and you'll go on till you'll go back by 50 terms right okay i don't need to write down what is the first term don't worry about it so assume that you have a arithmetic progression over here whose first term is 12 75 second term is 12 74 third term is 12 73 in other words the common difference is minus 1 okay so first term is 12 75 common difference is minus 1 number of terms is 50 how do you find the sum of such a series so the sum of such a series will be nothing but n by 2 to a n minus 1 into d okay so i don't have to find which is the first term of that set i don't need it whether you start from first to the last or last to the first it is an ap of course from first to the last it is the increasing ap last to the first it will be a decreasing ap but nevertheless it doesn't change our sum isn't it so this we can write it as 25 times this is 12 5 5 0 minus 49 so this is 25 times 2501 which is nothing but 6 25 25 well done arithra aditya good you have got the answer for this nice is it fine any questions any questions any concerns any questions no questions all set sir we thought we were good at this chapter sir but we are proven wrong see now you're already good at the chapter but now you're going to see different facets of problem solving involved right i agree with it i agree with you that it's an easy chapter but sometimes problems have become complicated okay let's take this question before you start solving a small correction i would like to make here this is a 4000 a1 a2 a3 till a 4001 is in arithmetic progression such that 1 by a1 a2 plus 1 by a2 a3 and so on till 1 by a4000 a4001 is equal to 10 and it is also given to us that a2 plus a4000 is equal to 50 find a mod of a1 minus a4001 and let's have around three three and a half minutes let me launch the poll for you time starts now all right two minutes gone i have got two responses so far very good one of the options is leading all right let's stop this in another uh 10 seconds five four three two one go okay option b has got the most number of votes okay out of 10 people who voted unfortunately only 10 eight of you have said option number b b for batinda what every time bangalore only batinda also okay now see let's use this information how do we even start that's what anurag wants to know okay anurag let's start this is given to us as 10 right okay now what i'll do is i'll multiply this with a d d being the common difference of this arithmetic progression so let d be the common difference of this arithmetic progression so i'll multiply this with d and i'll also multiply the numerators also with d d d okay d d d d d d okay now this d that you have the first d every d is different actually yeah so pay attention the first d is actually a 2 minus a 1 okay second d is actually a 3 minus a 2 okay third d is a 4 minus a 3 so as the number below it i'm basically converting my d in that particular way okay so this last d is a 4001 minus a 4000 okay up till here i don't think so anybody has any kind of a cesta or a problem okay if you have do let me know now first term over here can i write it as 1 by a 1 minus 1 by a 2 second term 1 by a 2 minus 1 by a 3 third term 1 by a 3 minus 1 by a 4 till 1 by a 4000 minus 1 by a 4001 a 4001 it is equal to 10 d but these terms will start cancelling okay ultimately what will happen everything will get cancelled except for 1 by a 1 and 1 by a 4001 and this is 10 d for us okay okay let's write this in the form of a simplified expression a 1 into a 4001 and numerator will be a 4001 minus a 1 that is equal to 10 d now we all know that a 4001 we all know that a 4001 is a 1 plus 4000 d isn't it because it's an arithmetic progression so a 4001 minus a 1 is 4000 d so numerator will become 4000 d so i can write the same thing as 4000 d by a 1 a 4001 as 10 d in other words a 1 a 4001 is equal to 400 let's call it as 1 now a 2 plus a 4000 is 50 now remember a 2 plus a 4000 is same as a 1 plus a 4001 so this will also be 50 let's call it as second equation right now what do i have to find out we have to find out this guy okay so we have to find out modulus of a 1 minus a 4001 okay let's say i call this as an x can i say x square is nothing but a 1 minus a 4001 square which is actually a 1 plus a 4001 square minus 4 a 1 a 4001 basically i use the formula a minus b whole square is a plus b whole square minus 4 ab and why did i choose to use this formula is because this term is known to us as 50 this term is known to us as 400 so 50 square minus 1600 will give you 900 so x square is 900 so x is 30 please do not write minus 30 because x is a positive quantity it's a positive quantity as you can see so only plus 30 will be the answer so option number b which most of you have answered with absolutely correct option is it clear any questions any questions any concerns here anything you would like to ask from this solution okay all fine all right so before we move any further we'll talk about properties of an arithmetic progression properties of an arithmetic progression uh the first property is the one which i've already discussed with you that the terms equidistant from the beginning and the end is same as the sum of the first and the last term and it is twice the middle term this is already discussed so just i'm mentioning it so that we don't forget about it second property is second property is if let's say a1 a2 a3 etc okay are in ap with common difference as d with common difference as d then please note the following then number one a1 plus or minus k a2 plus or minus k that means either you add or subtract a given constant k from these terms it will still be in an ap okay so if you have terms in an ap and you start subtracting a fixed number from all these terms let's say i start subtracting 2 2 2 2 from all the terms remember the resulting sequence will also be an arithmetic progression with common difference can anybody tell me what is the common difference will common difference change yes i hear an absolutely common difference will not change it will still be the same okay because what is common difference common difference is the difference of subsequent minus the preceding which is anyways not going to change because both are diminishing by the same or both are increasing by the same amount if you multiply if you multiply by a constant k then the resultant sequence will also be in ap with common difference now as kd with common difference as kd same goes with division as well okay if you divide by non-zero quantity this will still be in an ap with common difference as d by k okay but very important thing very very important thing if you raise it to a power of k let's say a1 to the power k a2 to the power k a3 to the power k please note this will not be in ap okay so 1 2 3 if i start raising it to a power of 2 so i get 1 4 9 this is not an ap because 1 4 common difference is 3 9 4 common difference is 5 okay if you start taking their log they will not be in ap if you start raising them to a constant they will not be in ap and so on okay anyways and let's say there is another ap b1 b2 b3 da da da da da with common difference as d dash with common difference as d dash any question anybody has sir is there any square progression uh progression see we definitely have a progression where there are squares of numbers but we cannot call that as arithmetic progression right arithmetic progression should have that basic characteristic that the difference of subsequent minus the proceeding should be fixed of course we have squares of you know terms available and you know how to find their sums out and we know their nth term but they will not be called ap's okay so here we are talking about them being ap's arithmetic progression okay now remember the fact that if you add it to ap's let's say a1 is added to b1 a2 is added to b2 a3 is added to b3 da da da da da this will still be an ap okay with common difference with common difference being d plus d dash okay similarly if you subtract them i'm doing it in the same way so that we don't have to write it that will be again in an ap with common difference as d minus d dash okay but however if you multiply them if you multiply them please note they will be not in ap okay same goes with if you divide them they will not be in ap so one ap divided by another or one ap terms divided by the terms of another ap will not be in ap okay please remember this okay uh i think second property we already did we are going to third property now so please allow me to change the slide if you want to copy it please do so yes akash reciprocal of arithmetic progression is what is geometric is a harmonic progression we'll talk about it when we do harmonic progressions none underdog another those who are copying dam sir underdog yes okay thank you let's move on to the next third property is the twice of terms so how do we how do we choose terms in ap especially when we know something about their sum okay now this is this is with a disclaimer that this is a preferred way of choosing the terms let me not uh brainwash you hear that it is the only way to choose a term no it is a preferred way to choose a term so if somebody asks you to choose three terms in ap so if you want to choose three terms or three numbers which are in ap then preferably you should choose it like a minus d a and a plus d okay it's not that if you choose it as a a plus d and a plus 2d your answer will not come nothing like that but if you choose it like this and you have an information about the sum of these numbers then you can easily find out a because as you can see when you add them the d vanishes right so 3a will be that sum given to you so your a can be easily be found out so especially where there is an information regarding the sum of the numbers you should always go or preferably go rather with a minus d a and a plus d as the numbers okay if you have to choose four terms in ap if you have to choose four terms in ap again you should preferably go with a minus 3d a minus d i'm sorry a minus d a plus d and a plus 3d as you can see here again also these terms are so written that they are balanced out so if you add them these will get cancelled off and the common difference is no longer d in this case the common difference here actually becomes 2d but d is just a name right don't be like so hardwired that d has to be a common difference no d is just a name which is used okay so here 2d is a common difference what big d correct if you have to choose five terms in ap preferably again word preferably you should use it you should choose a minus 2d a minus d a a plus d a plus 2d okay if you have to choose six terms in ap you should preferably go for a minus 5d a minus 3d a minus d a plus d a plus 3d a plus 5d okay I hope you have you can see the trend here the trend is these guys will follow each other and these guys will follow each other so let's say if you want to write down the seventh term just take the fifth term and write one term to the left one term to the right so seven terms will be created if you want to write eight terms take the sixth term like put one more term to the left one more term to the right so you'll have a minus 70 and here a plus 70 like that okay so this can go on and on let's take questions on this let's take questions very easy question I mean probably you would have seen this question millions of times in your schools also just a warm-up question the sum of three numbers in ap is minus three and their product is eight find the numbers I'm just going to give you 60 seconds for this not more than that it's a super easy question oh wonderful I've got the answers some people Kathy I don't know what is a what is d it is your call you are calling something as ad which is totally unaware to me so I just want the three numbers very good okay let's do this so again if you're asked to find a take three numbers which are in ap and no you know their sum okay then preferably take them as these numbers right so what are the benefit benefit is when you add them and equate it to minus three immediately you get the value as a which is minus one okay so you know your numbers are one minus d sorry minus one minus d minus one and minus one plus d now their product is eight so if you multiply them you'll get something of this nature so this is nothing but minus one square minus d square so which is one minus d square and this is eight so d square is equal to nine so d could be plus minus three okay so uh the there are two arithmetic progressions possible with this so if you take a plus three your answer would be minus four minus one and two and if you take a minus three your answer would be exactly four minus two minus one and minus four okay nevertheless this is not going to change the three numbers which are going to be still minus four minus one and two okay but if they ask how many ap's are possible okay then you have to say two ap's are possible in this case okay remember ap's need not always have d as positive you can have a negative d ap as well is this fine any questions any questions any concerns all right so before we move on to gp i would like to do one more question with you one more question yeah there you go and ap consists of even number of terms to n having middle terms so there are two middle terms of course because there are even number of terms so middle terms are one and seven respectively if n is the maximum value which satisfies this inequality find the value of the first term okay let's start the poll let's have around three three and a half minutes for this time starts now read the question properly before you start working it out very good one person has responded and what happened yeah i'm sure you would have done something wrong last one minute okay five four three two one go take a call i mean just take a guess if at all you are not able to solve it take a guess you should also check how accurate you are if you take a guesswork okay b is the most preferred guesswork which people take okay somehow b is the safe haven for people who want to take a guess most of you have gone with b by the way okay let's check whether b was correct or not see first of all if you have two n terms what are the positions of the middle terms so it is very obvious that the middle terms would have been t n and t n plus one am i right right let's say there were eight terms so the middle term will be the fourth and the fifth term isn't it correct yes or no let me write it see t1 t2 t3 t4 t5 t6 t7 t8 so these two will be the middle terms isn't it okay so if there are two n terms middle term will be nth and n plus one itself right so one thing is for sure that this is one and this is seven okay so t n plus t n plus one is known and therefore t1 plus t2 n will also be known and that is nothing but eight not only that the common difference is also known to us is six isn't it what is common difference common difference is n plus one at term minus n at term take any two consecutive term succeeding minus preceding is common difference so few things i came to know about this sequence or about this arithmetic progression number one uh some of the first and the last term which is t1 plus t2 n is going to be eight okay and common difference is six common difference is six okay so i can say one more thing here uh t2 n minus t1 would have been two n minus one into six correct anybody has a doubt with this information okay now i'm planning to use these two to generate t1 into into t2 one so you can see here there's a product of t1 into t2 t2 n so i'm planning to use one and sorry this is two one in two for that so what i'm going to do is i'm going to square them i'm going to square them and add them so if i square this i'll get t1 minus sorry tn minus not add them this subtract them subtraction makes more sense okay so square them subtract them that means you're doing 64 minus 36 to n minus one square if you see this this will give you four t1 t2 n am i right correct me if i'm wrong okay and this is we giving you 64 minus 36 to n minus one square drop the factor of four this will be 16 this will be nine okay now as per the question t1 t2 n plus 713 is greater than equal to zero that means 16 minus 9 to n minus one square plus 713 is greater than equal to zero what does it mean 729 minus 9 to n minus one whole square is greater than equal to zero drop the factor of nine 81 minus 2n minus one whole square is greater than equal to zero which is nothing but 2n minus one whole square minus 81 is less than equal to zero 81 could be written as nine square right so use a square minus b square formula so a square minus b square will become a plus b and a minus b correct if you want you can drop the factor of two from everywhere yeah so this means your n should lie between minus four to five i'm sure all of you know your inequalities we have done so much of problems so the question is saying that your n is the maximum value which satisfies this inequality means n is going to be five okay now this is basically an attempt by these people to make this question slightly more difficult okay this topic is already easy so how to make it difficult do all these inequalities add those right it'll become difficult right people will start you know making mistakes here so see j problems are like that only what they will do is they'll give you you know five different things from where you'll fetch the result to you know get the final answer so if you make a mistake in any one of those five things you are gone so as a result one problem will help them to test you on five skills or five concepts okay that's how j problems are normally thing so if n is five that means you know there are 10 terms okay there are 10 terms now my question is find t1 how will i do that okay let's find it out so i know t1 plus t10 is eight okay that means i know t1 plus t1 plus 9 into 6 is 8 that means 2 t1 is 8 minus 54 which is minus 46 so t1 value is minus 23 my dear chanta option d is correct not b okay how many people voted for d let me check only two people who are they don't answer me me me me sir i was the person let's be honest what was my theory oh people answering me yeah yeah anurag has a question so wouldn't be better if we add one into i did one in two oh see that also would have got but my problem was here i wanted to get my n value from this right so that's why i wanted to generate i wanted to generate your t1 into t2n in terms of n that was my concern here anurag if let's say that process helps you please go ahead and try to see whether you are able to find out i think you'll be stuck with t1 and n in the variables because one of the variables t1 or t2n will stay in the system no n will also be unknown t2n will also be unknown so how will you find you know one way where two variable inequalities from a single equation that would be difficult right and see what you're saying if you add them you get two t2n right two t2n plus 713 and on the right hand side you like end up getting a term in n so there'll be two variables involved t2n and n will be involved so single inequality will not help you to find that answer okay all right yes so we will now go on to geometric progressions i'm sure you would all like to take a break right now