 Hello, students. Good evening. Good evening to all of you. So, today we will do exercise five of sequence and series chapter. So, let's begin. Let's begin with question number one. Exercise five. So it is saying if the aim of two positive numbers, a and B is twice of their GM. Okay, so two numbers are there and B. So what will be the arithmetic mean of these two numbers, it will be a plus B upon two. And what will be the geometric mean, geometric mean of these two numbers will be under root AB. Now question is saying the arithmetic need is twice the geometric move. So, we can write it as a plus B upon two is equal to two times the root of AB. So, in both, both sides we will get a square plus B square plus two AB is equal to this will be four into four 16 AB. Right. So, a square plus B square minus 14 AB is equal to zero. So, divided by AB. So dividing by AB we will get a by B plus B by a minus 14 is equal to zero. Right. So, let us assume a by B is equal to x, assuming that we will have x plus one by x minus 14 is equal to zero. So it will become x square plus one minus 14 x is equal to zero. So it is a quadratic in x. Maybe write it as x square minus four x plus one is equal to zero. So, x is equal to minus being is 14 plus minus under B square 14 square is nothing but 196 minus four easy easy or one so it will be for one lead upon to it. Right. So, it will become 14 plus minus under root of 192 under root of 192 upon two. The root of 192 will be 64 into three. So, it will come outside of the root and root three will be as it is. So, it's this divided by two. So, we are having two values of x. The first value is x one, it will be seven plus four root three and extra will be seven minus four. And what is x, x is nothing but a by B. In the question, it is given that a is greater than B. Right. So if a is greater than B, if a is greater than B, then this ratio must be greater than one. But here, if you see seven minus four root three is less than one, hence, we will reject this value. This value will be a not accepted. And we have to find the ratio of age to be only so a is to be is equal to seven plus four root three. Right. We can rewrite it as four plus three plus four root three. So, it will become two squared plus root three squared plus two into two into root three. So it will become two plus root three whole square, right. So, we can further simplify it as two plus root three, two plus root three, and multiplying by its conjugate in numerator and denominator. We will get we will get two plus root three upon two minus root three into this thing will become a square minor we can apply the formula. This is a plus B into a minus B. So it will become a square minus B square B square is three. So, it will become one, right. This complete value will become two plus three upon two minus three. So, this will be our answer. So a is to be will be two plus three upon two minus three seconds option is correct. Option is correct for this question. Okay. So, let's move to the next question. Next question. If even a to G1 G2 and H1 H2 are two arithmetic geometric and harmonic means respectively between quantities a and B. Okay. So, there are two arithmetic means between a and B. There are two geometric means between a and B. And there are two harmonic means between a and B. So, these are arithmetic means these two are geometric means and these two are harmonic means. Then which of the following is not the value of AB, like we have to find the, okay, we will go through options and we will check whether that value is equal to AB or not. So, let's take the first option, a one into H2. So, what will be even what will be even we can write a one as to a must be upon three and what will be H2. What will be H2 H2 is the second harmonic mean between a and B, right. So, we can write it as three upon. It will be one by a plus two by the right. We used to write harmonic means in this way only. So, let's simplify it, it will become to a plus be upon three. So, if we are writing AB as LCM in that denominator this AB will move in the numerator and it will become B plus two way, right. So, here we can see this to a plus be will be cancelled out these three these three will be cancelled out so we are left with AB. So, even H2 is equal to AB, right, hence option a cannot be the answer, because we have to derive the value, which is not. Equal to AB. So, even H2 is equal to AB, hence this cannot be our answer. Let's check the second option. So, let's check out second option B. So, this is B is equal to a two into H1. So, what will be the second, what will be the value of a two a two is the second arithmetic mean between a and B. So, we can write a two as a plus to be upon three. Okay, and each one what will be the value of each one it is the first harmonic mean between a and B. So, AB will be equal to three upon upon a plus one upon. Okay. So, let's simplify it, it will be a plus to be upon three into three AB will go in the numerator and it will become to be plus a. Again, this a plus to be a plus to be will be cancelled this three and three will be cancelled out and it will be equal to AB. So, a to the product of a to and each one is also equal to AB, hence be will not be our answer. Now let's check option see G1 into G2. So, it is simple only it will be a AB, like a product of geometric means is equal to geometric mean of AB, like geometric mean of AB to the race to the power and right G1 G2 will be nothing but geometric mean of AB raised to the power and this will be AB raised to power N by two, right, and what is any of N is to only so it will be equal to AB. Hence, option see the product of G1 is G1 G2 is also coming to be AB, right. So, see also cannot be our answer. It will be D, like none of these none of the options given a correct. So, option D will be correct for this question. Let's go to the next one. Question number three. Now, what will be the geometric mean between minus nine and minus 16. The numbers are minus 16 and minus nine. And we have to find the value of geometric mean we have to insert geometric mean between these two numbers. So, what is geometric mean between two numbers a and B. It is under root AB right. The value of a and B, here we get minus 16 into minus nine. So 16 into nine is nothing but 144. So, it will become 12 right. Let me tell you one thing. This answer is wrong. This answer is wrong. The geometric mean of minus 16 and minus nine can never be a positive number. How can I mean like two numbers are given minus 16 and minus nine. So how can a mean of these two numbers be positive. The geometric mean of these two numbers must be negative, right. This must be negative. So, like, if the numbers are negative while finding geometric mean, if the numbers given are negative, like a is also negative and B is also negative. So, there is change in the formula that should be geometric mean should be minus of AB. Right. So this geometric mean will be minus of 12, not plus of 12. So, our answer will be minus 12. So option B is correct. This is clear to all like mean is something that will be in between the numbers. So that number must be between minus 16 and minus nine. It can never be positive. It can never be plus 12. It has to be minus 12. So, our answer will be minus 12 option B. Okay. So, let's move to the next question, question number four. So it is saying n belongs to natural number and n is greater than 25. If a g h denote the arithmetic mean geometric mean and harmonic mean of 25 and in. The least value of n for which this age age belongs to 2526 up to n like it this age is should lie between 25 to n and this age age must be a integer. So, for fulfilling this criteria what will be the least value of n. Okay. So, we are having two numbers, 25 and n. So, what will be the arithmetic mean. So arithmetic mean for these two numbers will be 25 plus n upon two. And geometric mean of these two numbers will be under root 25 and we can simply write it as five under root n. So, what will be harmonic mean harmonic mean will be two times a b means two times 25 and upon 25 plus n. Right. So, having given these two numbers we have found out the arithmetic mean geometric mean and harmonic mean. This, the value of a g and h should be integer, right, because as per given criteria it should lie between 25 to n. It should lie between 25, like it should be in between 25 and n, and we have to find the least value of n. So, for having for a to be integer n must be odd, right, it must be a odd number. And for having g to be integer, this n should be a perfect square, perfect square also. So, n is odd and n is perfect square so let's move to the options itself. So, 49 if we see 49 49 is a odd number 49 is a perfect square also. Similarly, 81 odd number plus perfect square plus one then 169 it is also odd number and it is a square of 13. The option is 225. Now, we can do one thing, we can put the options, like the value given in the options in this H, like harmonic mean, and we will see whether it is coming and integer or not. So, if you put the options. 225 this option D option D this, let me put first two into 25 into 225 upon 25 plus 225. This option this 225 will give me an integral value of H, which will be equal to two into, let's write it as 50 into 225 upon this will be 1550. We know five times this one 545. So putting n equal to 225 we are getting, we are getting age, the value of H is 45. Rest, you can put other options also like 49 81 and 169 the value of age will be not coming and integer. So, our option D is correct. So, we will take this as our answer option D. Now, we will take the next question, question number five, question number five. It is saying that nine harmonic means are inserted between two and three, right. So, to between two and three we are inserting nine harmonic means so H one H two H three up to H nine and three. So we are inserting nine harmonic means between two and three, then the value of a plus upon H plus five, where a is any arithmetic means and each is the corresponding harmonic. Okay, so, similarly, let's make one more sequence, which will be like, we will be putting nine harmonic we will be inserting nine arithmetic means, so it will be even a to a three up to a nine. And three, so these are nine arithmetic means between two and three. And we have to find the value of this expression. Okay, where it's like, if we take the second arithmetic mean the corresponding harmonic mean to be taken in this expression. So, we are taking a to like second arithmetic mean it should be second harmonic mean or else wise if you take the fifth arithmetic mean it should be fifth harmonic mean. So, let's check it out like, what will be the case. Okay, the harmonic mean. Let's generalize this. So, what will be the case arithmetic mean between two and three. B in minus K plus one. Into a plus K into B upon N plus one. Right. How it is coming. Suppose a and B are suppose a and B are two numbers and we are inserting in arithmetic means between these numbers. So, for finding Keith arithmetic mean, this is the formula where is the starting term and B is the last term, where is the first time we can say the is the first time and B is the last time. So, the key arithmetic mean will be given by this formula. So what is in here, we are inserting nine harmonic means so N is equal to nine. So, nine plus one will be 10 10 minus K. And what is a is the first number to us K into what is BB is three and N plus one will be nine plus one equal to 10 right. So, 20 minus 2k plus 3k will be plus K upon 10. So this will be the Kate Kate arithmetic mean between a and B. So what will be the key to harmonic mean between a and B. So, the key to harmonic mean will be in plus one upon N minus K plus one upon a plus K upon B. Where N is similar N is equal to nine and a and B are equal to two and three respectively. So N plus one will be 10 upon 10. Sorry, yeah, N plus one again 10 minus K. What is a is to. And what is be be is three. So we can simply write it as. 10, taking this LCM six will go in the numerator, and we will be left out with three into 10 minus K plus 2k. This will be nothing but 60 upon 30 minus 3k plus 2k will be 30 minus K, right. So this will be the gate. Harmonic mean. Okay, now we will go to this expression, what is asked in the question a plus six upon a plus six upon H plus five. Okay, so let us consider the Kate arithmetic mean and the Kate harmonic mean right. So we have to find the value of this. Now, what is the case arithmetic mean, it is equal to 20 plus K upon 10. And this will be six upon what is the Kate harmonic mean Kate harmonic mean is 60 upon 30 minus case or 30 minus K. Let me raise it. So it will be six upon 60 upon 30 minus case or 30 minus cave in the numerator plus five right. So, let's cut it and it will be 10 taking LCM we are having 20 plus K plus 30 minus K plus. This will be 50. So this plus K minus K is getting cancelled out. We are left with 100 upon 10. So, let's see the actually this expression is not dependent on the value of K, like it is not dependent on the position, because this Kate Kate term is getting cancelled out from the, from our expression right. So, it will be equal to 10 for any value of K you can take. Right. So, our option, our answer will be 10 to this question. Let's see which option is there. So, it will be option C. So option C will be correct. Moving ahead. Let's take the next question is saying if each one is to an H and B in harmonic means between a and B. Okay. So a and B are two numbers and we are inserting in harmonic means between it in between a and B. Right. So, we have to find the value of this expression h one plus a upon h one minus a plus. And it's harmonic mean plus B upon any time on it mean minus B. Okay. So, let's simplify it like, let's first find out the value of a first harmonic mean and the nth harmonic mean. So, what will be the first harmonic mean, first harmonic mean will be, we are inserting in harmonic means so it will be in plus one upon in upon a plus one upon the right. So, we can simplify it as n plus one into AB upon B and plus a right. And what will be our nth harmonic mean, or in a harmonic mean will be in plus one upon one by a plus in upon the. In plus one into AB by B plus a and. Okay. Now let's try to solve this expression. So, each one plus a each one and writing as in plus one AB upon a plus B and this is each one plus a upon. In plus one AB upon a plus being minus a right. We move a bit right hand side. And what is this. This is the nth harmonic mean, which is n plus one AB upon B plus a and must be upon n plus one AB upon B plus a and minus be right. So, this will become n plus one AB plus a into a plus being. And it will become n plus one AB minus a into a plus being, and this denominator term a plus being will be answered. Similarly, let's really rewrite this as n plus one AB plus B into B plus a and and this will be n plus one AB minus B into B plus a. Okay. So, what can we do further simplifying further we get. Let's open the bracket. It will be easier than AB and plus AB plus a square plus AB and on it will become AB and plus AB minus a square. And minus AB and similarly this will become AB and plus AB plus V squared plus AB and and then the denominator it will be AB and plus AB minus B squared and minus AB and. Okay. AB and minus AB and will be cancelled. This similarly here also it will be cancelled. So, we are left with AB plus a square plus two times AB and and in denominator we have. Let's take a common so it will be B minus a. Okay. So, we have two times AB and plus AB plus B squared, right. And in denominator if we take B as B as common it will be a minus B. And here we are having a into B minus a. Okay, so let's take negative outside from this bracket it will be. It will be minus B minus and this is okay so let me change this. It will become minus here and it will become B minus a. And it will become B minus a here and negative sign we come here. So, taking LCM AB into B minus a. So, it will be multiplied with B here. Right, so multiplying by B, we will have AB square plus a is square B. Plus to AB is square and minus what will be this thing this numerator thing will be multiplied by a. And it will be to a is square B and plus a is square B plus a B is where. Okay. So, moving further. This AB is square and this minus AB square will be cancelled out. Then, similarly this a plus a square B this minus a square B will be cancelled out. We are left with to AB square and minus to a square B and upon AB into B minus a. So, in numerator we can take to AB and format to AB and format so we will have to be minus a upon AB into B minus a. So, this B minus a this B minus a will cancel out this AB and this AB will be cancelled out so this expression the final value will be to N. So, let's see whether it is there in the option or not to N. So, yeah, it is there option C is correct. Option C is correct for this question. So, let's move ahead to question number seven. It is saying the aim of to given positive numbers is to. Okay. And if the larger number is increased by one, the GM of the numbers becomes equal to aim of the given numbers. Okay. So, what will be arithmetic mean of these two numbers it will be a plus B by two. And it is given to be two. So arithmetic mean is two for these two numbers so we can write it as a plus B is equal to four. Let's make it as a question one. If the larger number is increased by one, suppose I'm taking B is greater than a. Okay. So now numbers will become a comma B plus one. So if the larger number is increased by one, the GM of the numbers. Sorry. So what will be the GM of these two numbers. The GM will be under root a into B plus one. So the GM of the numbers becomes equal to the arithmetic mean. So, as per question, this under root a into D plus one becomes equal to the arithmetic mean so what was the arithmetic mean arithmetic mean was to. Equal to two. So squaring both sides, we will get a into B plus one is equal to four. So it will be a B plus a is equal to four. So let's take it as equation two. Now, a plus B equal to four and a B plus a equal to four so we can write it as a plus B equal to a B plus a. This a, we can cancel it out. So we are left with a B is equal to be so a is equal to a is equal to one right. If a is equal to one, our B will be three. Since a plus B is equal to four. So we got the value of a we got the value of B. And now question is asking the harmonic mean of a and B, right. So what will be harmonic mean of these two numbers. It will be two AB upon a plus B. So two times a is one B is three. What is a plus B a plus B is four. So, this is equal to three by two. So harmonic mean of these two numbers will be three by two. So option is correct. Now, question number eight is the last question of this exercise. So let's check it out. If a a one a two a three dot dot dot a two and B are in AP. Then again one sequence is given and they are saying it is in GP and it is the harmonic mean of a and B. Then what will be the value of this series. Okay. It is given a a one a two a three up to a two in and be this is in AP. Okay. And a be one be two be three be two and a and a term and this be isn't GP right so we have to find the value of this and one more thing is given H is the harmonic mean. H is the harmonic mean of a and B so we can write H as two AB upon a plus B. Now let's see what what we have to find out what we need to calculate. It is a one plus a two and upon B one into B two and plus a two plus a two and minus one. That will be B two into B and minus one plus dot dot dot dot up to a and plus a and plus one upon B and into B and plus one. So, if we can see this a one plus a two and like. The question is asking for these terms and this term, these terms, so a one is the second term from starting and a two and is the second term from the last in this AP. So, we know that the sum of numbers if we distant from the starting and to the end is equal to some of first number and last number. So, we can write this as a plus B, right. And this be one into B to win. So, question is asking for this product, this be one into B to win, and this sequence isn't GP. So here also we know that the product of terms from starting equidistant product like equidistant terms from starting and equidistant from and the product of these two numbers is equal to AB, like the product of first and last number. So, it will be AB and same thing, this a two and a two and minus one this a two this is the second term sorry third term from starting and where will this a two and minus one and will lie, it will be the third term from it. So, it will be also equal to a plus three. And similarly this between to be two and minus one will be equal to AB. Now, how many terms are there, like, if you observe this is starting from a one and it is going up to N. So, total N terms are there, a plus B upon N, total N, N terms are there. So, it will become N times a plus B upon AB. Okay, now, what is a plus B upon AB from here we can write a plus B upon a BS to upon H, right. So, it will be in into a plus B by AB is equal to two upon H. So, this will be two and upon H. Option A is correct. Option A is correct for this question. So, we have finished this exercise. And I think it is not that much tough. Okay, we are closing this and we will meet you soon. Okay, Tata, till then goodbye, take care.