 Hello everyone. We are once again back in our video series where we are providing the solutions to the questions given in the exercises of Arihan's book. So we have covered the two exercises of sequence and series chapter. So today we are going to start the next exercise, exercise number three. So let's proceed. So here comes the exercise three, question number one. The fourth, seventh and last term of GPR, 1080 and 2560 respectively. The first term and the number of terms in GPR. Okay. So let's proceed with the given information. It is given that the fourth term is 10. Okay. So a fourth term is 10. The seventh term is 80 and the last term or we can say the nth term is 2560. Okay. So since this series is in GP, let's consider it as AAR, AR square up to AR raise to power n minus 1. So we can consider these terms in GP. So as per given information, the fourth term is 10. So what will be the fourth term? The fourth term will be AR cube. AR raise to power 3 is equal to 10. And the seventh term, the seventh term will be A into AR raise to power 6 is equal to 80. So substituting the value of R cube in this second equation, let's consider this as equation one and this as equation two. So we can rewrite the equation two as A into R cube whole squared is equal to 80. So A into R cube can be replaced by 10 by A. 10 by A square is equal to 80. From here, we will get the value of A. So this will be A into 100 by A square is equal to 80 or 1A will be cancelled out and A will be 100 upon 80. That is nothing but 5 by 4. So we got the value of A. What is A? A is the first term of this GP. So since we got the value of A, we can easily derive the value of R also. So put the value of A in equation one. It will be 5 by 4 R cube is equal to 10. From here, we will get R cube is equal to 10 into 4 by 5. So this is nothing but 8. So R comes out to be 2. So we have find out the value of A and R. Now it's turned to find the value of N. So use this information. Our nth term is equal to 2560. So nth term of a GP is given by A R raised to power N minus 1 and this is equal to 2560. So put the value of A and R in this. So this will be 5 by 4 2 raised to power N minus 1 is equal to 2560. It will become 2 raised to power N minus 1 equal to 2560 into 4 upon 5. Cancel out this 512. So 2 raised to power N minus 1 is equal to 42048. So this 2048 can be written as 2 raised to power 11. So N minus 1 is equal to 11. From here we get the value of N as 12. So total number of terms in the given progression is 12. So the first term and the number of terms. So the first term is 5 by 4 and the number of terms is 12. So option number C is correct. Okay. So let's move forward to the next question. What is our next question? Okay. Question number 2 is saying that if the first and the nth terms of a GP are A and B respectively, and if P is the product of the first N terms, then P is square is equal to. Okay. So it is saying that P is the product of first N terms. So let's consider the GP as A, A, R, A, R is square, A, R, Q, dot, dot, dot up to. Nth term that will be A, R raised to power, sorry, A, R raised to power N minus 1. So this is our product P. Now the question is asking to find the value of P is square. Okay. So if you see this P can be simply written as A raised to power N since there are total N terms and R raised to power 1 plus 2 plus 3 up to N minus 1. Right. So this can be further simplified as A raised to power N into R. What will be the sum? 1 plus 2 plus 3 up to N minus 1. This will be nothing but N, N plus 1 by 2. So it will become this minus 1 plus 1 will be cancelled out. This will become N into N minus 1 upon 2. So R raised to power N, N minus 1 upon 2. So this is our P. Now the question is asking to find the value of P square. So let's square it on both sides. It will become A raised to power 2 N into R raised to power N into N minus 1. Now we can further rewrite as A raised to power N, sorry, A is square into R raised to power N minus 1 whole raised to power N. Right. So further we can split this A square as A into A raised to power N minus 1 whole raised to power N. And this can be written as A into this A raised to power N can be written as B because the Nth term is given as B. So this will become A into B raised to power N. So this will be our value of P square. So this is our answer. A B raised to power N. So option C is correct. So this question is done. Let's move to the next question. Question number three. It is saying if A1, A2, A3 are three successive terms of a GP with common ratio R, the value of R for which A3 is greater than 4A2 minus 3A1. Okay. So A3 should be greater than 4A2 minus 3A1. Okay. And A1, A2, A3 are the three successive terms of a GP. Okay. So A3 can be written as A1 into R square. This should be greater than 4 into A2. A2 can be written as A1 into R minus 3A1. Now we can divide this left hand side and both sides. We can divide both sides by A1 since A1 is positive. Okay. And the equality sign will remain same. So divide both sides by A1. It will become R square is greater than 4R minus 3. Let's take this thing to the left hand side. It will become R square minus 4R plus 3 greater than 0. So this is quadratic in R. And we have to find the value of R which will satisfy this inequality. So let's factorize this. Let's factorize this quadratic equation. So it will become R square minus 3R minus R plus 3 greater than 0. Taking R common from the first two terms, it will become R minus 3 minus 1 R minus 3 greater than 0. So R minus 1 into R minus 3 greater than 0. Now we have to find the solution shape for this. So this is an inequality. So first let's draw the critical point. So critical points are 1 and 3. So for the values R greater than 3, this whole thing will be positive. For the value between 1 and 3, it will be negative. And for R less than 1, both these factors will become negative and overall it will be positive. So the value of R ranges from minus infinity to 1 union 3 to infinity. Or the same thing can be written as R less than 1 or R greater than 3. So this is our solution shape. R less than 1 or R is greater than 3. So option C is correct for this question. This is clear to everyone. So let's take the next question. Question number 4. Question number 4 is saying if x2x plus 2, 3x plus 3 are in GP, then the fourth term is looking simple only. x2x plus 2 and 3x plus 3. These terms are in GP. So we have to find the next term or fourth term. So since these terms are in GP, so 2x plus 2 upon x should be equal to 3x plus 3 upon 2x plus 2. This is nothing but common ratio. We can write common ratio as 2x plus 2 upon x. We can write common ratio as 3x plus 3 upon 2x plus 2 and both these are equal. So cross multiplying we will get 2x plus 2 whole square is equal to x into 3x plus 3. So let's open it. This will become 4x square plus 4 plus 8x is equal to 3x square 3x square plus 3x. So let's take everything on one side. It will become x square plus 8x minus 3x it will become 5x plus 5x plus 4 is equal to 0. So we can further split the middle term and factorize it. It will become x square plus 4x plus 6 plus 4 is equal to 0. Taking x common it will become x plus 4 plus 1x plus 4 is equal to 0. So x plus 1 into x plus 4 is equal to 0. From here we are getting the two values of x. x is equal to minus 1 or x is equal to minus 4. So these two values we are getting. x is equal to minus 1 or x equal to minus 4. But what the question is asking? The question is asking for the fourth term. So let's take x equal to minus 1. When x is minus 1 what will these terms become? It will become minus 1 comma 2 into minus 1 minus 2 plus 2 is 0. And the third term will become 3 into minus 3 plus 3 is 0. So actually this is not in GP because in GP no terms can be equal to 0. So x equal to minus 1 is rejected. x equal to minus 1 is rejected. Now let's take x is equal to minus 4. So taking x is minus 4 our terms will become minus 4 comma 2 into minus 4 will be minus 8 plus 2 that will become minus 6. 3 into minus 4 will be minus 12 plus 3 means minus 9. This is in GP. So now we have to find the fourth term. So what will be our fourth term? So let's find r first. So r will be minus 6 upon minus 4. This will become 3 upon 2. So our fourth term will be r into third term or this 3 by 2 into third term or third term is minus 9. So this will become minus 27 upon 2. This is nothing but minus 13.5. So this is our fourth term minus 13.5. So option D is correct. Option D is correct for this question. So it's done. Now we can move ahead to the next question, question number 5. This is the question number 5. It is saying in a sequence of 21 terms, the first 11 terms are in AP with common difference 2. And the last 11 terms are in GP with the common ratio 2. If the middle term of AP is equal to the middle term of GP, the middle term of the entire sequence is okay. So let's create one sequence that is A1, A2, A3 up to A11. Then this will go up to 21 terms. So our 21th term. So this is our sequence. Now the question is saying that the first 11 terms are in AP. So these first 11 terms are in AP with common difference of 2. And it is also stating that the last 11 terms. So last 11 terms means what? It will start from A11 and it will go up to A21. So these last 11 terms are in GP and for which the common ratio is 2. So this is the given information. And one more thing is given the middle term of this AP, the middle term of the AP is equal to the middle term of GP. So what will be the middle term of this AP? Like total terms are 11. I am considering only this first 11 terms of AP. So total terms are 11. So our middle term will be 6th one. Our middle term will be 6th one. This will be the middle term for this AP. And how can we represent this 6th term? 6th term can be represented as A1 plus 5D. And what is D for this AP? D is 2 for this AP. So A6, we can write A6 as A1 plus 10. This will be our 6th term or the middle term of the AP. Now what will be the middle term of this GP? Look this A11, this 11th term will be actually the first term of the GP. This 11th term will be the first term of the GP. And similarly here also the total number of terms are 11. So the 16th term. So 16th term will be the middle term of AP. Like this middle term of GP. This will be the 16th term. So A16. And how can we write A16? This we can write as A11 is the starting term. And the common ratio, common ratio R raised to power 5. So our 16th, A16 or the middle term of GP will be equal to A11. And R is given to, so it will be into 2 raised to power 5. So this will be our middle term of GP. Now question is saying that both these terms are equal. Both these terms are equal. So we can equate this. So equating this, both these terms we get A1 plus 10 is equal to A11 into 2 raised to power 5. We can write it as 32. Okay. Now can we find the value of this A11? Yeah, we can find because this is the last term of the AP. This is the last term of the AP. So we can write this A11 in terms of A1 so that we can find the value of A1. So this will become A1 plus 10 is equal to and what will be the A11? A11 will be A1 plus 10d. 10d means what? 20. So A1 plus 20 plus 32. Am I doing something wrong or what? Okay, okay, okay. This will not be plus, this will be multiplied, patient side. Right. So this A1 will be written as A1 plus 20. Multiply it by 32. Multiply it by 32. So we get A1 plus 10 is equal to 32 A1 plus 32 into 2, 64, 640. From here we get the value of A1 as this will become 31A. Let it write completely 31A is equal to 10 minus 640 that will be minus 630. Okay, from here we get A1 as minus 630 upon 31. So this is our first term. This is our first term. Now we got the first term of the sequence. Now what we have to find? We have to find the middle term of the entire sequence. Okay. So for entire sequence, the middle term will be this A11. Right. Middle term of the entire sequence will be A11 and A11 is nothing but A1 plus 10 times of D. So A1 we got this is minus 630 upon 31 plus 10 D is nothing but 20. So simplify it. It will become 31 this minus 630 plus 31 into 2 will be 62. So this will become 620. So finally we get minus 10 upon 31. So this will be A11 and A11 is the middle term of the entire sequence. So this will be our final answer minus 10 upon 31 option A option A is correct. Okay. So we are done with this question also. Let's have a look to the next question. Question number six. It is saying three distinct numbers x, y, z form a GP. In the order that the number 7x plus 5y, 7y plus 5z, 7z plus 5x form an AP. The common ratio of GP is okay. So it is saying that x, y, z form a GP. So let's write it x, y, z is in GP. And these numbers 7x plus 5y, 7x plus 5y comma 7y plus 5z, 7y plus 5z comma 7z plus 5x. These terms are in AP. Now we have to find the value of common ratio of this GP. Okay. So since these three terms are in AP, we can write it as 7y plus 5z minus 7x minus 5y. This is the value of D. We can write it as D equal to this. And this is nothing but third term minus second term. So this will be 7z plus 5x minus 7y minus 5z. So from here we get 7y minus 5y, it will become 2y. 2y plus 5z minus 7x is equal to 7z minus 5z will be 2z plus 5x minus 7y. So take all the terms to one side, we will get this 2y plus 7y, it will become 9y, 5z minus 2z. It will become 3 plus of 3z then minus 7x minus 5x that will become minus 12x is equal to 0. Now it is divisible by 3 I think, so we can write it as 3y plus z minus 4x is equal to 0. So what we have to find actually, we have to find the value of common ratio of GP. Okay. So what will be common ratio of this GP? Common ratio of this GP will be we can write it as y upon x or we can write it as z upon y. This will be the common ratio for this GP. Okay. So let's divide this equation by y. So it will become 3 plus z upon y minus 4 times x upon y, this will be equal to 0. Now z upon y is nothing but our common ratio that is r, this is nothing but r. And what is this x by y? This x by y is actually 1 by r, right? I am taking this to this side. 3 plus 2 upon z upon y is r, so 3 plus r minus 4 by r is equal to 0. So let's solve it. It will become 3r plus r square minus 4 is equal to 0. Or 3 square, sorry r square plus 3r minus 4 is equal to 0. Further we can simplify it as plus 4r minus r minus 4 equal to 0. Taking r common r plus 4 minus 1 r plus 4 equal to 0. So r minus 1 multiplied by r plus 4 is equal to 0. From here we are having two values of r, r is equal to 1 or r is equal to minus 4. r equal to 1 or r equal to minus 4. So there is no one in the option. So we will go with the option a that is minus 4. So minus 4 is correct answer for this. So this question is done. Now let's move to the next one. Question number 7. The sum of in terms of the series. One series is given here 11 plus 103 plus 1005 plus dot dot dot. This is going up to infinity. No, this is going up to n terms. And we have to find the this sum of n terms. Okay. So this 11 can be written as 10 plus 1. And this 103 can be written as 10 square plus 3. Similarly this can 1005 can be written as 10 cube plus 5. Now this series will go up to n terms in the similar fashion. So let's write it separately. So we can write it as 1 plus 3 plus 5 plus 7 up to n terms and 10 plus 10 square plus 10 cube plus 10 raise to power 4. This is also going up to n terms. So we know the value of the first n odd numbers that is equal to I think n square. Okay. Let's we can find the value also. So n terms. So what will be the sum? It will be n by 2 2 times of first term to a plus n minus 1 into d d is 2 here. And what will be this? This is a GB with common ratio 10 and the first term as first term as 10 first term as 10 and the common ratio is 10. So we can write this sum as a raise to power n minus 1 upon r minus 1. Okay. So this will become n square. I know because this plus 2 minus 2 will get cancelled out. This will become 2 n in the bracket. It will become 2 n. 2 n square upon 2 plus 10 into 10 raise to power n minus 1 upon 9. So this will be our answer. 10 by 9 10 raise to power n minus 1 plus n square. Yeah. Option C is correct. So let's move to the next question. Question number 8. Increasing GB the sum of the first and the last term is 66. The product of the second and the last but one means the product of the second and second last one is 128. And the sum of the terms is 126. So let's collect the information first. What it is given in the question. The sum of the first and the last term. So let's assume a GP first. Our traditional GP a a r a r is square up to a r raise to power n minus 1. Okay. So we have considered total n terms. So the sum of the first. And the last term like this a plus a r raise to power n minus 1 is equal to 66. This is the first information we got the product of the second and the last but one. I think this mean second last only. So the product of the second. What is the second term? This is a r into second last one. What will be the second last one term? That will be a r raise to power n minus 2. So a r into a r raise to power n minus 2 is given as 128. This is the second information we got. And what is the third information? The sum of the terms is 126. So what is the sum of this GP? Sum of this GP can be given as a into r raise to power n minus 1 upon r minus 1. This is given to be 126. So these three informations we are provided with. What are the unknowns here? If we see a, r and n. So three unknowns, three equations. So definitely we can get a solution from this set of informations. But how to retrieve it? Let's check it out. Let's go ahead with this equation too. Let's go ahead with this equation. So from equation 2 we can see a square. We are rewriting this equation too. So a into a a square r raise to power n minus 2 into r. This will become r raise to power n minus 1. And this is equal to 128. So from here we can get r raise to power n minus 1 as 128 upon a square. Now we can use this thing. We can use this value of r raise to power n minus 1 in the equation 1. So putting this value in equation 1, what we will get? Whether we have write it, wrote it correctly or not? The sum of the first term and the last term. The first term is a, last term is a, r raise to power n minus 1. That is 66. Okay, let's try it out. So this will become putting the value of this in equation 1. We will get a plus a into 128 upon a square is equal to 66. It's working. So a one times a will be cancelled out. It will become a square a square plus 128 is equal to 66 a. So let's take this 66 a to the left hand side. It will become a square minus 66 a plus 128 is equal to zero. So we can split this middle term 64 into two. Yeah. So a square minus 64 a minus 64 a minus two a plus 128 is equal to zero. Take a common from here. It will become a minus 64 take minus two common from here. It will become a minus 64 is equal to zero. A minus two into a minus 64 is equal to zero. From here we are getting the two values of a either a is equal to two or a is equal to 64. These two values we are getting is equal to two or a is equal to 64. Okay. So let's check it. So we know from here we know our race to power in minus one. So our race to power in minus one is equal to 128 upon a square. Now when a is equal to 64 we will get our race to power in minus one is equal to 128 upon a square. That means 64 into 64. This will go two times this will become one upon 32. And if we take a is equal to two this our race to power in minus one will be equal to 128 upon two into two. So four three two. So our race to power in minus one will become 32 when when a is equal to two and when a is equal to 64 it is becoming one upon 32. But one thing we have to notice here it is an increasing GP right. The question is saying it is an increasing GP. So if it is increasing GP or must be greater than one. But from here we are getting are less than one when we are putting a equal to 64. So this implies that R is less than one. So this option is we have to reject this one. So A is two. This is confirmed that A is two. This is the accepted value. Now if A is equal to two we are having our race to power in minus one is 32. What are the other informations we are provided with? We have not utilized this third equation yet. So let's take this third equation in the room. So third equation is saying A into our race to power in minus one upon R minus one is equal to 126. Okay. So we can write. So put this A equal to two here. We will get two into our race to power in our race to power in can be written as from this we can write our race to power in upon R. This will become our race to power in minus one. This is equal to 32. R race to power in can be written as we can write our race to power in is 32 R. So replace our race to power in by 32 R minus one upon R minus one is equal to 126. We are getting something. Okay. From here we can find the value of R. So this will become 63. So 32 R minus one is equal to 63 R minus 63. Right. So it will become 63 R minus 32. 32 42 52 62 31 R. 31 R is equal to 63 minus one 62. So R from here we get RS two. So we got the value of A. We got the value of R also is also equal to two and R is also equal to two. Now what else we have to find? We have to find the value of N. Right. The number of terms in the series. So let's utilize this equation. R race to power N minus one. From here if you see, let me go downward. Okay. I'm changing the color of the pin. R race to power N minus one is equal to 128 upon A square. Now we know the value of R and A. So put that it will become two raised to power N minus one is equal to 128 upon A square. A square means four. So this will become four, three, two. So two raised to power N minus one is equal to 32. We can write it as two raised to power five or N minus one is equal to five. So N equal to six. So total number of terms in the given progression is six. This will be our answer N equal to six. Whether yeah, option A, it's given in the option A. So option A is correct for this. So this question was a bit lengthy, but okay, we have done it. Let's move to the next question. This is board number eight. So I have to go to board number nine. Okay. Here we are. If S1, S2, S3 be respectively the sum of N, 2N and 3N terms of a GP, then the value of this expression S1 into S3 minus S2 divided by S2 minus H1 whole squared. Okay. So what is S1? S1 is sum up to N terms. Okay. What is S2? S2 is sum up to 2N terms and S3 is sum up to 3N terms. So we have to find the value of this S1 multiplied by S3 minus S2 upon S2 minus H1 whole squared. Okay. I think this question, mechanical type, let's say, let's calculate. We have to do work on this S1, S1 sum of N terms. So sum of N terms of a GP, we can write it as A into r raised to the power N minus 1 upon r minus 1. This is sum up to N terms. What will be the S3? S3 sum up to 3N terms. So it will be, A is the starting term of the GP. Okay. So it will be common. So A into r, r will be also common because this is the same GP. So r raised to power, only difference will be in the power of r. That is r raised to power N, total number of terms. It will be 3N here. r raised to power 3N minus 1 upon r minus 1. And similarly, A into r raised to power 2N minus 1 upon r minus 1. And what is there in the denominator? It is S2 means A into r raised to power 2N minus 1 upon r minus 1 minus S1. A into r raised to power N minus 1 upon r minus 1. And this, total car, fully square. Right? Okay. So do, we can do one thing. A into r raised to power N minus 1 upon r minus 1. Now, let's take A upon r minus 1 common from these two terms. Okay. So this will become A upon r minus 1. And in bracket, it will be r raised to power 3N minus 1 minus of r raised to power 2N plus 1. Right? Similarly, take A by r minus 1 common from this. So it will become r raised to power 2N minus 1 minus r raised to power N plus 1. And this will be a square. So we can see here, these things will be cancelled. This A by r minus 1, this is nothing but A square by r minus 1 whole square. So these things got cancelled. So we are left out with, we are left out with r raised to power N minus 1. Here, this minus 1 plus 1 will be cancelled. Similarly, this minus 1 plus 1 will be cancelled. This will become r raised to power 3N minus r raised to power 2N divided by r raised to power 2N minus r raised to power N. So we can do one thing. We can take r raised to power 2N common from this bracket. And from denominator, we can take r raised to power N common. So let's take it r raised to power N minus 1. If we take r raised to power 2N common from here, it will become r raised to power N minus 1. Right? And if we take r raised to power N common from here, it will become r raised to power N minus 1. Yeah, whether I have done something wrong, wrong, wrong. Okay. These things, this square will come. So if we take r raised to power N common here, it will be 2N and this will be square. So r raised to power N minus 1, r raised to power N minus 1, 2 times in the numerator. This will be cancelled and this r raised to power 2N and this r raised to power 2N will also be cancelled. So this will become 1 finally. So it was earlier looking a bit so much what you say. Difficult, but finally we got rid of it and we got answer as 1. So option A, option A is correct for this. So we are done with this question number 9. Let's move ahead with question number 10. So question number 10, if mod A is less than 1 and mod B is less than 1, then the sum of series. Okay, one series is given and we have to find the sum of that. So series is 1 plus 1 plus A multiplied by B plus 1 plus A plus A square multiplied by B square plus 1 plus A plus A square plus A cube multiplied by B cube and this is going up to infinity. Right, because dot, dot, dot is in continued fashion. So we have to find the sum of this series. So what can we do? Let's open the bracket first and see what we are getting. So this will become 1 plus B plus AB plus B square plus AB square plus A square B square plus B cube plus AB cube plus A square B cube plus A cube B cube plus dot, dot, dot. So whether we are getting anything convenient. So 1 plus B plus B square. Okay, let's write it as 1 plus. Now take the terms having only B. So we will get this B. We will get this one we have already taken. So this B, B square, B cube and further it will be B4 also. So B plus B square plus B cube plus dot, dot, dot it will go up to infinity. Now take the terms which are having A. So if we take that then this we will cover AB, AB square then AB cube. Okay, so this will become B plus B square plus B cube plus dot, dot, dot up to infinity. Now similarly take the terms having A square. So A square we are having this thing, this thing A square then this thing. Okay, so first term is having B square then B cube and it will continue in the similar fashion. Similarly, the terms with A cube will also come. And seeing the pattern I can realize that it will be having the terms B cube plus B4 plus B raised to power 5 within the bracket. Similarly, A raised to power 4 will be having B raised to power 4 plus B raised to power 5 plus B raised to power 6 and this series will go on. Right, so what can we do further? This is the GP and we are given that mod B and mod A are less than 1. This is the sum of GP for infinite terms. It will become the sum of GP up to infinite terms A upon 1 minus R. Okay, so what is A here? Means first term is B and 1 minus R and the common ratio is also B. So this we got from here. From here what we will get? This is also some up to infinity terms. The common ratio is this. So this first term is B and common ratio is also B. So this will be 1 minus B. Similarly this A square and B square upon 1 minus B plus A cube into B cube upon 1 minus B plus dot dot dot this will go. So what we are getting now? So this will be 1 plus B upon 1 minus B, let it be as it is. I think I am able to see something. In all these terms I am taking 1 upon 1 minus B common. So what will be remaining terms? What will remain in the packet? It will be AB plus A square B square plus A cube B cube plus this will continue up to infinite terms. Right? So 1 plus B upon 1 minus B plus 1 upon 1 minus B and we can write this sum of infinite terms of this GP as AB upon 1 minus what is the common ratio here? AB. So first simplifying this take 1 minus B and 1 minus AB as LCM. So this will become 1 minus B into 1 minus AB plus B into 1 minus AB plus AB. So let's open the packet. It will be 1 minus AB minus B plus AB square plus B minus AB square plus AB upon 1 minus B into 1 minus AB. So minus B plus B got cancelled AB square minus AB square minus AB plus AB cancelled. So we are left with 1 upon 1 minus B into 1 minus AB whether this is there in the option or not. Let's check it out. 1 upon 1 minus B 1 minus AB 1 minus B 1 minus AB yeah. It is there. Option C is correct. Option C is correct for discussion. So I don't know I approached through this method. If anyone have some alternative or any shortcut method please come to me and like we can discuss on that. So let's close this question here only. So let's move to the next one. Question number 11. Okay. It is saying if the sites of a triangle are in GP and its larger angle is twice the smallest then the common ratio are satisfies the inequality. Okay. So. Triangle is there. So triangles have three sites and for considering three numbers in GP we prefer taking A by R, A and A by R. So here we assume that A by R like this. Let's assume this side as A by R. This side as AR and this as A. So we can do that and what then the common ratio are satisfies the inequality. Okay. Let's consider A by R as the smallest side smallest side and this AR as the largest side. So since A by R is the smallest. So angle opposite to it will be less. Right. So if I'm taking this as theta and angle opposite to this AR since this is the largest side we have considered. So it will become two theta. Right. Its larger angle is twice the smallest. So it will become two theta. Now we have to find the value of R. We can apply sign rule for this triangle. That is A by R upon sin theta is equal to AR upon sin two theta. So this will become A by R sin two theta is equal to AR into sin theta. So this I'm cancelling this A with A it will become sorry. So R square will become sin two theta upon sin theta. Now we can write sin two theta as two sin theta cos theta upon sin theta. So cancelling this we get R square is equal to two cos theta. So we got R square as two times of cos theta. Now the maximum value of cos theta can be one. But cos theta we get cos theta one at theta is equal to zero degree. So theta cannot be equal to zero degree since it is angle of a triangle. So theta cannot be equal to zero degree. So R square should be always less than two. So R must be less than root two. So R maximum we can say R maximum must be less than root two. Now while considering this right while considering this sides of the triangle R must be greater than one. Why? Because if suppose we consider R less than one suppose we take R is equal to half. So what will be our sides? Our sides will become A upon one by two A and A into one upon two. That is two A A and A by two. But we have assumed that this is the smallest side but now it is becoming like largest side. So our assumption will be wrong. So hence R must be greater than one. So where should R lie? R should lie between one and root two. So this will be the range of R. This will be the range of R. R must be within R must be greater than one but less than root two. So option B is correct. Option B is correct for this question. So we are done with this. Let's take the next question. It is saying if A x cube plus B x square plus C x plus D is divisible by A x square plus C then A, B, C, D are in A, B, G, P or HP or none of these. Okay. So it is saying that A x cube plus B x square plus C x plus D it is divisible by A x square plus C. Right? Divisible means what? It is completely divisible. Means remainder will be zero in this case. So let's divide it by our long division method. Take x. So it will become A x cube plus C x. So plus C x. Now let's subtract it. It will become zero. B x is super will come as it is. It will become zero. And B x square plus D will remain. Now what we can do? We can take it as B upon A. B upon A. Right? Then only A is it. So B upon A. So B upon A into x square will be, let me write this side. B upon A into A x square plus B upon A into C. So it will become B x square, B x square. This A will be cancelled and B C upon A plus B C upon A. Now let's subtract it. This will be cancelled. It will become D minus B C upon A. Now this is the remainder, right? But the remainder should be zero here. Because question is saying this, that this cubic polynomial is divisible by this A x square plus C. So since this remainder must be equal to zero, we can write it as D minus B C upon A is equal to zero. Or A D minus B C is equal to zero. Or we can say A D is equal to B C. So suppose we are having four terms A B C D. So let's take these are in GP. Okay? So B upon A must be equal to C upon B and that must be equal to D upon C. So okay. So this cross multiplying we get A D is equal to B C. So since A D equal to B C means these A B C D are in GP. These are in GP. So this is the answer. So option B is correct. This result we got, right? A D equal to B C. Now we have to check it whether it is in AP or GP or HP. So I just gone with this assuming that these are in GP and we got the correct relation as we derived from the required condition. So these are in GP. A B C D are in GP. So moving to the next question, question number 13. It is saying if R N denotes the RRR up to N digits where R is equal to 1, 2, 3, up to 9. And A is equal to 6 suffix N then B equal to 8. N C equal to 4 to N. So R N denotes number R. This will be written up to N times, like N digits are there. Okay. So we can write 6 N as 6, 6, 6, 6, how many times? N times, like N digits. We have to write this 6 N times. So this will be our 6 N and this is equal to A. As per question, this is equal to A. Okay. And what is, and it is also saying B is equal to 8 N and C is equal to 4. But in foot it is written 2 N. So 2 N times it will repeat. This B will repeat, this 8 will repeat N times as it was repeating in 6. But C will repeat 2 N times. Okay. First let's solve this A. Then I think we will get some idea. So take 6 common form here. Like then things under the bracket, I mean within the bracket can be written like this in digits. So 6, this number can be like, we can rewrite this number in the bracket as 10 raised to power 0 plus 10 raised to power 1 plus 10 raised to power 2. How many times? 10 raised to power N minus 1. Because total, how many digits are there? N digits. So it's starting from 0, 10 raised to power 0. So it will end, so it will end at N minus 1, 10 raised to power N minus 1. So this is GP actually, right? So this is GP having 1 as the first term and 10 as the common ratio. So 10 raised to power N minus 1 upon 10 raised to power N minus A, R raised to power N minus 1 upon R minus 1. R is 10, 10 minus 1. So this is our A. Let us simplify it. So it will be 6 into 10 raised to power N minus 1 upon 9. Okay. So I think in the similar fashion we can find the value of B also, right? So B will also become, B will become instead of 6, it will come 8 and rest all things are similar only. So 8 into 10 raised to power N minus 1 upon 9. So this will be the value of our B, right? But in C, C the digits are repeating up to 2 in terms, right? So we can write C as 4 into, the only difference will be instead of this N, like 2 N will come, right? So 10 raised to power 2 N minus 1 upon 9. Rest all things will be same only. So this will be the value of B and this will be the value of C. Okay. Now what the question is asking? Then A square plus B plus C equal to 0 or A square plus B minus A equal to 0, A square plus B or A square. Like everywhere in every option A square B is there. So let us first calculate it. A square plus B thing. Like this A square plus B because it is common to every option, no? So let us check it out. So what will be A square? A square will be 36 into 10 raised to power N minus 1 upon 9 is squared. So this will be A square and B. B is 8 into 10 raised to power N minus 1 upon 9, okay? So we can take 4 into 10 raised to power N minus 1 upon 9 common. We can take this common. So what will we left out with? 4 into 9, right? So this will be 9 and this thing is squared so it will come one time. 10 raised to power N minus 1 upon 9. And plus 8. So we have taken it will be 2, okay? I think I am doing right. So 10 raised to power N minus 1, okay? So this 9 and 9 will be cancelled out. We will have 4 into 10 raised to power N minus 1 upon 9 into this will be 10 raised to power N minus 1 plus 2. This will be plus 1. So it will be 4 into this A minus B A plus B. So A square minus B square we can write it as. So A square like 10 raised to power 2 N A square minus B square minus 1. The things in the numerator I have written here and 9 will be as it is. So I think this was the value of C which we have. Yeah, it is there. Value of C 4 into 10 raised to power 2 N minus 1 upon C. So A square plus B is equal to C. This is equal to C. Means A square plus B is equal to C. This is the relation between A, B and C. This is our answer. A square plus B equal to C. A square plus B is equal to C. A square plus B equal to C. So option B. Option B is correct. A square plus B minus A equal to 0 or A square plus B equal to C. So we are almost close to this, to the end of this chapter. So 0.427 recurring. 0.427. 27 are repeating terms. Represent the rational number. Okay. So suppose this is the value of X. This type of problems we used to do in like class 10. In first chapter, I think real numbers. So let's multiply it by 10. Why multiply it by 10? Because non-recurring figures are only one figure is non-recurring. So I'm multiplying by 10. So it will become 4.27. And now multiply by 100. Let's suppose it as equation 1. Now multiply it by 100. So it will become 1000 X is equal to 427.27. And consider it as equation 2. Now subtract equation 1 from equation 2. It will become 990 X is equal to. This will become 427.27 minus 4.27. So this will become 424. So 990 X is equal to 424. From here we get X is equal to 424. This will be 423, you know. 423. So 423 upon 990. I think it is divisible by 9. 9. 4 times 36. 6 into 3. That will be 747 upon 110. So option B is correct. Now coming to the question number 15. This is the last question of this exercise. So we are at the end. Let's do it quickly. If the product of three numbers in GP be 216. So three numbers in GP. So we prefer considering number as A by R, A and AR. So the product of these numbers are 216 and their sum is 19. So let's take the product A by R into AR is equal to 216. So AR will be cancelled out. It will become A cube is equal to 216. So A we will have as. So A is 6. Now their sum is 19. So let's take the sum A upon R plus A plus AR is equal to 19. So A is 6. So sorry. 6 by R plus 6 plus 6R is equal to 19. So taking R as LCM we will get 6 plus 6R plus 6R is equal to 19R. So it will become 6R is square plus 6R minus 19R it will become minus 13R plus 6 is equal to 0. So we can break it as 9 and 4. So 6R is where minus 9R minus 4R plus 6 is equal to 0. Take three are common. So first two terms it will be 2R minus 3 6R is where minus 9R. Okay. And take minus 2 common from here it will be minus 2. This will be 2R minus 3 is equal to 0. So 3R minus 2 into 2R minus 3 is equal to 0. So from here we get R is equal to 2 by 3 or R is equal to 3 by 2. These two values of R we are getting. So what it is asking then the numbers are. So okay when R is equal to 2 by 3 our GP will become A means 6 upon R 2 into 3 comma 6 comma A into R. R is 2 by 3. So it will be 9 comma 6 comma 4. Okay. And when R is equal to 3 by 2 I think it will be opposite to it. Let's check it. Our GP will become that means terms of the GP will be A 6 upon R is 3 into 2 comma 6 into 6 into 3 by 2. So it will be 4 comma 6 comma 9. Yeah 4 comma 6 comma 9. So the numbers are 4, 6 and 9. Option A is correct. So we are done with this exercise. Actually the number of questions in this exercise was 15. So it took a long time but we tried to summarize it like close it as early as possible. So anyhow the length of the video will be more but can't help with that. So okay we are closing this session right now and we will meet again. Okay. Bye.