 So, hello everyone hope you all are doing well. So, today we will be solving exercise 4 of chapter sequence and series. So, till now we have solved the 3 exercise 1, 2 and 3 exercise where we have dealt with sequence problems problems based on AP and GP. So, this exercise mainly consists of mixed problems on the arithmetic progression geometric progressions and harmonic progressions. So, let's begin let's begin with this exercise. So, let's have a look at question number 1. It's saying that A, B, C are in AP and B, C, D are in HP then following options are given and we have to choose the correct one. So, A, B, C are in AP and B, C, D are in HP. So, these two things we are provided with. So, since A, B, C are in AP we can write 2B is equal to A plus C and we can write here C as 2 times BD upon B plus C since B, C, D are in harmonic progressions. So, let's take it as equation 1 this as equation 2. Now, I will be substituting the value of C in the equation 1. So, it will be 2B is equal to A plus C, C in place of C I will write 2BD upon B plus D. Sorry, it will be B plus D here. So, taking LCM it will be AD plus AB plus 2 times BD is equal to 2B. Now cross multiply it we will get 2B square plus 2BD is equal to AD plus AB plus 2BD. Now these 2BD terms will be cancelled out from this equation and we will be left with 2B square is equal to I am taking A common here. So, it will be B plus D. So, 2B square is equal to A into B plus D in place of B plus D we can write 2BD upon C, right. So, 2, 2 will be cancelled out 1B will be cancelled and we will be left with BC is equal to AD. So, AD is equal to BC, option B is correct. So, this will be our answer AD is equal to BC. So, let us move to the next question, question number 2. If ABC are in AB then A by BC, A by C and 2 by D are in. Okay, so, ABC these 3 terms are in AB and we have to say A upon BC comma A upon C comma 2 upon B. Like these terms will lie in AB, GP or HP we have to see. So, since ABC are in AB we can write it as 2B is equal to A plus C. Now, let us check what is asked in the question it is saying A by BC. So, if I divide this equation by BC on both hand sides I will be getting A upon BC term. So, let us try it out. So, 2B upon BC is equal to A upon BC plus C upon BC, right. So, this B will be cancelled out it will be 2 upon C is equal to A upon BC plus this C will be cancelled. 1 upon B. So, if the question was like in place of this 2 by B if the question was like A upon sorry 1 upon B. So, we can easily say that these terms will lie in AP. Because 2 times of middle term is equal to sum of A first term and third term like if second term was in AP then it will be the 2 times of second term will be equal to first term plus third term like what we did here, right. But the question is saying it's 2 by B it's not 1 by B. So, none of these will be answered to this question because none of the answers is matching, right. So, either there is some problem in the question itself like it should be 1 upon B if the if it was in AP but okay let us assume the question is correct. Our answer will be none of these, right. So, let's have a look at the next question question number three. So, if ABC are in AP and ABD are in GP, then A comma A minus B comma D minus C will be in. So, ABC are in AP and ABD. So, ABD is in GP. Now, we have to say A comma A minus B comma D minus C will be in AP, GP or HP. So, from here we get to B is equal to A plus C, okay. Let us assume this as equation 1 and ABD are in GP. So, B square will be equal to AD. Let's take it as equation 2. Now, how should we approach like we will try to have these terms in any of the equation. Then only we can say whether these terms will be in AP, GP or HP, right. So, what I will be doing I will multiply this equation 1 by A. So, what by multiplying A we will have 2 AB is equal to A square plus AC, right. This can be rewritten as A square minus 2 AB is equal to minus of AC. I am able to say like A square minus 2 AB. If I add B square to this side then I will have A minus B whole square, right. So, let's do that. A square minus 2 AB plus B square since B square is not there in our earlier equation. So, I have to add B square on the right hand side also. So, it will be B square minus AC, okay. So, here I am getting A minus B whole square is equal to B square minus AC. B square minus AC and we know B square is equal to AD. So, we will replace this B square by AD. So, this A minus B whole square is equal to AD minus AC. A minus B whole square taking A common from here. It will be A into D minus C. Now, see this A A minus B and D minus C. So, middle term is squared is equal to product of the first and third term. Like this is the property of GP, right. So, A minus B whole square is equal to product of A and A minus C. Sorry, D minus C. So, these three terms are in GP, right. These three terms are in GP. So, option number B is correct. Let's move to the next question, question number 4. If X1 Z are in AP and X2 Z are in GP, then X comma 4 comma Z will be in, okay. So, X comma 1 comma Z, these three terms are in AP and X comma 2 comma Z, these three terms are in GP. Then we have to show like this X comma 4 comma Z will be in either AP, GP or HP. So, from here we can say 2 is equal to X plus 0, right. Let's take it as equation 1 and since X2 and Z are in GP, we can say 2 squared is equal to X into Z or 4 is equal to X into Z. Let's take it as equation 2. Now, we have to comment on X4 and Z, right. X4 and Z, whether it will be AP, GP or HP. So, let's assume this an AP. So, if these three terms were in AP, then 2 times 4 will be equal to X plus Z, right. So, A is equal to X plus Z, but we know X plus Z is equal to 2. So, this assumption is wrong, like X comma 4 comma Z can never be in AP. Let's try it out with GP. So, if X comma 4 comma Z is in GP, then 4 square must be equal to XZ or 16 equal to XZ. But we know that XZ is equal to 4 from our equation to from our given conditions. So, this can never, this can never be in GP also. So, only one option is left. Let's assume it has an, let's assume these three terms are in HP, right. So, if these three terms are in HP, then 4 must be equal to 2 times XZ upon X plus Z. So, let's see whether it is satisfying or not. So, 2 into XZ, we know XZ is 4 and X plus Z is 2. So, 2, 2 will be cancelled out, 4 is equal to 4. So, these three terms X, 4 and Z are in HP. Sorry. So, option number C, option C is correct for this question, right. So, we are done with this. Let's move ahead. Question number 5. So, it is saying if A, B, C are in GP, A, B, C are in GP, A comma A minus B comma C minus A comma B minus C are in HP. Then A plus 4B plus C, we have to find the value of A plus 4B plus C. Okay. So, A, B, C, it's given to be in GP. And so, from here we can say B square is equal to AC, right. B square is equal to AC. And the next information is A minus B comma C minus A comma B minus C are in HP. So, if these three terms are in HP. So, a reciprocal of these are must line AP, right. Or we can say, we can proceed directly also. Like C minus A, C minus A must be equal to 2 times A minus B into B minus C upon A minus B plus B minus C, right. So, this minus B plus B will be cancelled out and we will be left with C minus A is equal to 2 times AB minus AC minus B square plus BC. And denominator we will have A minus C, okay. So, let's take a minus sign common from it. So, it will be and I will be cross multiplying. So, it will be C minus A whole square is equal to 2 times AB minus AC minus B square B square is equal to AC, right. So, let me write it AB minus B square can be replaced by AC. So, it will be minus 2 AC plus BC. And since we have taken negative common from this denominator, I have to change the sign also. So, it will become minus AB plus 2 AC and this plus BC will become minus BC, right. So, this will be our final. So, this will be C minus A whole square is equal to 2 times 2 AC minus AB minus BC. So, let's open the square it will be C square plus A square minus 2 AC is equal to 4 times AC minus 2 times AB minus 2 times BC. So, from here we can do one thing I'm able to say this A square term C square term here 2 AB 2 BC. So, let's try to make out like make this equation in terms of A plus B plus C whole square. So, it will be A square plus C square minus 2 AC I'm taking all the terms to the left hand side. It will be then minus 4 AC minus 2 AB minus 2 BC is equal to 0, okay. So, further my approach is to make this equation in terms of A plus B plus C whole square. So, what extra we need here? We need sorry A square plus B square plus C square B square was not here. So, I will be subtracting it minus B square then this will be oh I've done one mistake this will be minus 4 AC this will be plus 2 AB plus 2 BC right. So, minus B square minus 2 AC minus 2 AC and minus 4 AC it will become minus 6 AC minus 6 AC plus 2 AB plus 2 BC for formula we need plus 2 CA also then I have to subtract 2 CA. So, 2 AB 2 BC is already covered now I am adding 2 CA additionally so I am subtracting it and this will be equal to 0. So, it will become A plus B plus C whole square right. So, these three terms this 2 AB 2 BC 2 CA is already covered. So, we are left out with minus B square minus 8 AC this is equal to 0. So, A plus B plus C whole square is equal to B square plus 8 AC right. Now, we know B square is equal to AC since ABC are in GB. So, it will become 9 AC right. So, A plus B plus C whole square sorry I am removing this square I am taking square root. So, this will be removed. So, A plus B plus C will be equal to plus minus 3 under root AC under root AC means what under root AC means B sorry. So, we can replace here only like 8 AC can be replaced with 8 B square ok. So, let me replace it by 9 AC can be written as 9 B square already here also we could have done this thing. So, it will be plus minus 3 B right plus minus 3 B A plus B plus C will be equal to plus minus 3 B. So, when I am taking this plus sign. So, it will be A plus B plus C is equal to 3 B or A plus B plus C is equal to minus 3 B. So, from this equation we will be getting out desired result like A plus 4 B plus C this was asked in the question. So, A plus 4 B plus C will be equal to 0. This is what we need and let this equation be like A minus 2 B plus C is equal to 0. So, either A minus 2 B plus C is equal to 0 or A plus no not both the equations are justified because A plus B plus C is equal to plus minus 3 B. So, when we take minus 3 B it will give A plus 4 B plus C equal to 0 and if we take a plus 3 B it will be A minus 2 B plus C equal to 0. So, 0 is the correct answer for this question. So, option A is correct. Let us move to the next question, question number 6. It is saying that the if the M plus 1th, N plus 1th and R plus 1th term of an EB are in GB and M, N, R are in HP then the value of the ratio of the common difference to the first term of AB. So, M plus 1th term, M plus 1th term of AP and N plus 1th term of AP and R plus 1th term of AP. These three terms of AP are in GB right. So, what will be the M plus 1th term of AP? It will be suppose I am taking A as the first term of the AP and D as the common difference. So, M plus 1th term can be written as A plus MD right and N plus 1th term we can write it as A plus MD and R plus 1th term will be A plus RD. Now, these three terms, these three terms are in GB, these three terms are in GB. So, it must satisfy the property of GB that is A plus MD whole square is equal to A plus MD into A plus RD right. Now, what is asked in the question? Then the value of the ratio of the common difference to the first term of the AP. Okay. So, question is asking to find the ratio of D is to A, where D is the common difference of this AP and A is the first term of this AP. So, let's simplify this equation, we will have A square plus N square D square plus two times of A and D is equal to A square plus A RD plus A MD plus M RD square right. So, this A square A square will be cancelled out. This will be left with N square D square plus two A and D is equal to let's take a common from these two terms or AD both. So, let's take AD common from these two terms, we will have M plus R within the bracket. So, AD into M plus R plus M RD square. Okay. So, it will become N square D square plus two A and D is equal to AD. Now, one more information is given in the question M in R in HP, right. We have not utilized this information yet. So, if M in R in HP, so we can write in as two M R upon M plus R right from here we will get M plus R is equal to two M R upon N. So, we will use this information. So, M plus R can be written as two M R upon N. And this will remain same M R B square right. So, what it will become from, from the right hand side, we can take again M R common, right. So, if we take M R common again, we will have two AD upon N plus D square. It will be M R into taking LCM it will be two AD plus N D square. And in left hand side, we will, we are having N square D square plus two AD. Okay. So, from here also we can take N common. So, I'm seeing a, what do you say? If we take N common from this, I'm seeing one common term on both hand sides. So, if we take N common here, it will be N D square plus two AD, right. So, let's bring this thing to the left hand side. We will have N into N D square plus two AD minus M R to AD plus N D. N D square upon N is equal to zero. So, from this we can take N D square plus two AD common. So, N D square plus two AD. If we take common from both these terms, we will be left with N minus M R upon N. So, N D square plus two AD into this thing will be equal to zero. So, from here we can say either N D square plus two AD is equal to zero or we can say N minus M R upon N is equal to zero. So, from here N is equal to M R upon N or N square is equal to M R. But N M R was in HP as per the question, right. N M R on N M R for M N R in HP. So, since this thing is there M N R in HP since M N R in HP so N square can never be equal to M R. If this M N R would have been in GP then only this thing would have been valid, right. So, this thing is rejected N square is equal to M R. From here we get N D square is equal to minus two AD. So, one D will be cancelled from both sides. We will have P upon A is equal to minus two upon N. So, this was asked in the question the ratio of the common difference to the first term. So, it will be minus two upon N. So, let's check whether this is there in option or not D by eight minus two by N. Yes, this option A is correct for this question. Option A is correct. So, let's take the next question, question number seven. It is saying that if ABC are in AP and A square B square C square are in HP then three options are given. We have to say which one is correct. So, ABC are in AP, ABC are in AP, A square B square and C square are in HP. So, from here we can say two B is equal to A plus C. Let's take it as equation one. And from here we can say since A square B square C square are in HP we can say B square is equal to two times A square C square upon A square plus C square. This is our equation number two. So, substitute the value of B here. So, it will be A plus C upon two whole squared is equal to two times A square C square upon A square plus C square. So, let's simplify it. It will be A square plus C square plus two AC upon four is equal to two A square C square upon A square plus C square. So, here also A square plus C square is there and here also A square plus C square is there. So, for sake of convenience let's take A square plus C square is equal to X. I am taking A square plus X is C square is X. So, it will become X plus two AC into X is equal to eight A square C square. So, it will become X square plus two AC X minus eight A square C square is equal to zero. So, it's a quadratic in X. So, we can split this middle term as four minus two four minus two eight. Okay. So, it can be rewritten as X square plus four AC X minus two AC X minus eight A square C square is equal to zero. What is this? It will be A square C square eight. So, minus eight A square C square is equal to zero. I have split this middle term. Now, take X common from first two terms it will be X plus four AC take minus two AC common here. It will be X plus four AC is equal to zero. So, X plus four AC into X minus two AC is equal to zero. So, from here we can say either X plus four AC is equal to zero or X minus two AC is equal to zero. Now, what was X? X was A square plus C square. So, A square plus C square plus four AC is equal to zero or A square plus C square minus two AC is equal to zero. From here we can see A minus C whole square is equal to zero. So, A minus C is equal to zero or we can say A is equal to C. You can say A is equal to C, but ABC was in AP as per the question. ABC are in AP, right? So, since ABC are in AP, so two B is equal to A plus C and A is equal to C. So, two B is equal to two C. So, B will be equal to C. So, from here we can say A is equal to B is equal to C, right? And from here, this was also one of the option, no? Either X plus four AC will be zero or X minus two AC will be zero. From here we get A is equal to B is equal to C. From here we can say A square plus C square minus two AC means plus two AC plus two AC from right hand side it will be minus two AC. So, A plus C square is equal to minus two AC. But nothing, this sort of nothing we are saying in the option. So, we are having only A is equal to B is equal to C in the option. So, yeah, A is equal to B is equal to C. So, this option is correct for this question. So, we are done with this question. Let's take the another one, question number eight. So, it is saying if ABC are in HP. ABC are in HP. Then A upon B plus C comma B upon C plus A comma C upon A plus B. We have to say whether these three terms are in A, B, HP or GB. That we have to identify. And the given information is ABC are in HP, right? So, since ABC are in HP we can say B is equal to two times AC upon A plus C. Okay, so how to approach this question? A upon B plus C, B upon C plus AC upon A plus B. So, we have to proceed like we have to check separately, like let's assume these three terms are in AP. If these three terms are in AP, then two times middle term that will be two B upon C plus A must be equal to A upon B plus C plus C upon A plus B, right? So, from here we will get B B plus C into A plus B. So, it will become A square plus AB plus BC plus C square is equal to two B upon C plus A. Now, I think it will be B squared from here, B cube. Like, we can, anyhow we can see, we can so if we can so like B is equal to two AC upon A plus C, then we can say that these terms are in AP. But from here I am not able to like simplify this. Anyway, it will be simplified in terms of B, but it will be a difficult one. So, let's hold this and let's check whether these things are in GP or not. So, if these three terms would have been GP, then B upon C plus A whole square must be equal to A upon B plus C into C upon A plus B, right? So, it will be B square upon C plus A square is equal to AC upon A plus B into B plus C. From here also I am not able to like get whether this will result in having the value of B or not. So, let's check with HP also. So, if these three terms were in HP, then B plus C upon A and C plus A upon B and A plus B upon C, these three terms must be in AP, right? So, applying the property of AP here, two times C plus A upon B must be equal to B plus C upon A plus A plus B upon C, right? So, it will be equal to let's take LCM and solve it. So, it will be BC plus C square plus A square plus AB, right? And this is equal to two times C plus A upon B, right? And we have not utilized this information that B is equal to two AC upon A plus C. So, here we can use like this A plus C thing. From here we know A plus C is equal to two AC upon B. So, we will utilize here. So, two upon B into A plus C. A plus C can be written as two AC upon B, right? And this will be equal to BC A square plus C square plus, let's take B common here. So, B common, after taking B common, we will be left with A plus C, right? Upon AC. So, let me go down what, let this thing be in terms of A plus C only. I think that will be more useful to us. Two upon B or two upon B and A plus C, right? Or let me check what it results in A square plus C square plus, no, I am doing something wrong, I think. So, this will be two upon B, C plus A can be written as like ABC varying HP, no? So, this will be B is equal to two times AC upon A plus C, right? Whether we are getting any fruitful here, A square plus C square plus B into A plus C upon AC, right? So, C plus A, two upon B. So, this will be B squared. B is equal to, okay, what like, let me erase it first. Anyhow, we have to prove like B is equal to two AC upon A plus C, right? If we can do that, then we can say like these three terms are in HP. So, let's erase this also. So, we can write it as what was here? B plus C upon A plus A plus B upon C, okay? So, two upon B into A plus C is equal to AC means this will become BC plus C squared plus A squared plus A squared plus AB, right? So, let's check the value of B, what will be the value of B from here? So, it will be or two by B is equal into A, sorry, two by B can be written as two by B, we can write two by B as A squared plus C squared plus, take B common from here, it will be A plus C upon AC into A plus C, right? So, it will be A squared plus C squared plus B. Now, A plus C, we can write it as two AC upon B, right? And this is equal to, this is equal to AC into A plus C. A plus C can be again written as two AC upon B, but for A plus C, okay? So, this B, this B will be cancelled out, this will be A plus C whole squared in the numerator, right? So, A plus C whole squared upon AC into A plus C. So, this one term will be cancelled, only in numerator, only A plus C will be left out. So, this will be equal to two upon B, or we can write as B is equal to two times AC upon A plus C. So, hence, like, we arrived the value of B, assuming that these three terms were in, assuming that these three terms were in HP, right? This was justified since we got the value of B as two AC upon A plus C. So, these three terms are in HP basically. These three terms are in HP, right? So, there is no alternative. We have to, like, go one by one only. We have to say whether these three terms are in AP or in GP or in HP. So, finally, we got that these three terms are in HP. So, let's move ahead to the next question, question number nine. Question number nine is saying that if X plus Y by two comma Y comma Y plus Z by two are in HP, then XYZ are in. So, this X plus Y upon two comma Y comma Y plus Z upon two are in HP, okay? Then we have to identify XYZ will be in AP, GP or HP. So, since these three terms are in HP, we can say that Y is equal to two times X plus Y upon two into Y plus Z upon two upon X plus Y upon two plus Y plus Z upon two, right? So, we can rewrite as this two, these two will be cancelled out. So, it will be XY plus XZ plus Y squared plus YZ upon this two X plus Y plus Y plus Z and this is equal to Y. So, this one by two, this two will be cancelled. We will have Y is equal to XY plus XZ plus Y squared plus YZ upon X plus two Y plus Z. So, cross multiplying after cross multiplying, we will have XY plus two Y squared plus YZ is equal to XY plus XZ plus Y squared plus YZ. So, this XY, this XY, this YZ, this YZ will be cancelled. So, two Y squared is equal to XZ plus Y squared or Y squared is equal to XZ. So, this XYZ are in GP. So, we can say XYZ are in GP. Since Y squared is equal to XZ. So, option B is correct for this question. So, one more question is left in this exercise. Question number 10. So, if A plus B upon A minus AB comma B comma B plus C upon 1 minus BC are in AP, then A comma 1 by B comma C are in, okay. So, it is saying A plus B upon 1 minus AB comma B comma B plus C upon 1 minus BC are in AP, right. So, two times of B will be equal to A plus B upon 1 minus AB plus B plus C upon 1 minus BC. So, simplifying this, we will have 1 minus AB into 1 minus BC in the denominator. And in numerator, we will have A minus AB C plus B minus B square C plus B plus, sorry, this will be B minus AB square. Plus C minus ABC, this is equal to 2B, right. So, this will be 2B into M cross multiplied. So, 1 minus BC minus AB plus AB square C is equal to what is written there, 1A plus 2B. So, this term I have taken, BB is taken, plus C, this is also taken, minus 2ABC, this is done, this is done, minus 2ABC minus B square C, right. So, open this bracket, it will be 2B minus 2B square C minus 2AB square plus 2AB cube C is equal to A plus 2B plus C minus 2AB C minus B square C, right. So, this 2B, this 2B will be cancelled out and this will be 2AB cube C is equal to what we are having in right hand side, A plus C minus 2AB C minus B square C, this will be plus 2B square C. And this will be plus 2AB square plus 2AB square. So, this minus B square C plus 2B square C, this will become plus B square C. So, I am erasing this minus term plus B square C plus 2AB square, ok. Now, I will take this 2AB C to this hand side, so to the left hand side. So, 2AB cube C plus 2AB C will be equal to A plus C plus B square C plus 2AB square, right. So, taking 2AB C common from here, we will have B square plus 1 is equal to A plus C plus, if we take B square common from here, we will have C plus, whether I have done something wrong, 2AB square minus AB square, ok, ok. I have left this term or what, minus 2AB C, ok, it is there, minus B square C and one more term will come here, no, minus B square C and minus AB square, right. So, minus AB square will also come here. So, minus AB square. So, it will become, if we take 2A is minus AB square. So, 2AB square minus AB square will be AB square. So, these two will be omitted actually. So, these two will be omitted. So, it will be AB square. So, B square I am taking common. So, it will be having what will be left out C plus A, right. So, 2AB C into B square plus 1 is equal to, again I am taking A plus C common here. So, it will have, we will be left with 1 plus B square, right. So, this 1 plus B square term, we can cancel it out. So, we will be left with 2AB C is equal to A plus C. So, what was question? What was the question asked? So, A comma 1 by B comma C, right. A comma, A comma 1 by B comma C. So, if these three terms were in HP. So, 1 by B must be equal to 2 times AC upon A plus V, right. Suppose, we will taken this in HP. So, from here if we divided by AC, we will have 2AB C upon AC is equal to A plus C upon AC. So, it will be 2B is equal to 1, sorry, 1 upon C plus 1 upon A. Or from here we can say directly that this AC AC will be cancelled out 2B is equal to 2B is equal to A plus C upon AC, right. So, this or B is equal to we can say A plus C upon 2AC. Same we are getting from here also B is equal to A plus C upon AC. So, these three terms A comma 1 by B comma C will be in HP, right. I am omitting this. So, this is all. So, let me mark the correct option. A comma 1 by B comma C should be in HP. This should be in HP. So, option C is correct. So, we are done with this exercise. Hope it is clear to all. We got stuck in some questions, but nothing we can do right now, like this is the nature of the question itself. We have to see whether this the ASP terms are in AP, GP or HP. So, this is all for today. So, we will be coming back once again with the next exercise, exercise number 5. So, till then, okay. Till then, goodbye and good night to all. Thank you.