 So, hello everyone. My name is Satish Upadhyay and we are going to start a video series like where we will be providing solutions to the exercises of Arihant's book. So, in this particular video series we will cover all the questions from the Arihant's book. And it will be quite helpful for those students who are like, who are practicing from this book. And I would like to suggest like, before going to see these videos, you first give a genuine, genuine try to attempt the questions given in the exercises so that it will be, it will value, it will add some value to your preparations. And like, you will feel more confident for attempting the questions. So, today's chapter is a sequence and series and we are taking exercise number one, like exercise for season one, station one. And this, we will be going to solve the, this particular exercise. Okay. So, let's start. Let's start with the first question. What it is saying, the first term of a sequence is one. And n plus one-th term is obtained by adding n plus one to the n-th term for all natural numbers n. So, the sixth term of the sequence is, okay. So, it's like, it has been given the first term of sequence is one. Okay. The first term of sequence is one. And it is saying like the n plus one-th term, our n plus n-th term is obtained by adding n plus one to the n-th term. What does it mean? Our n plus one-th term is obtained by adding n plus one to the n-th term. So, this is what the question is trying to tell us. Okay. So, from here, we can write n plus one-th term minus n-th term is equal to n plus one, right. And the question is asking to find the sixth term. Okay. So, if you put values of n here, like if we put n equal to one in this equation, what we will get? We will get t2 minus t1 is equal to one plus one, that is two. So, similarly, n is equal to two. When we put n equal to two, we will get t3 minus t2 is equal to two plus one-three. When we put n equal to three, it will give t4 minus t3 is equal to four. When n equal to four, it will give t5 minus t4 is equal to five. And finally, on putting n equal to five, we will get t6 minus t5 is equal to six. So, adding all these equations, what we will get? You can see like many terms are getting cancelled. Like this t2 will be cancelled by this t2, t3, t3, t4 minus t4, t5 minus t5. So, finally, we get t6 minus t1 is equal to two plus three plus four plus five plus six. Okay. And we know the value of t1. Even in the question, t1 is equal to one. So, putting the value of t1, what we will get? We will get t6 as one plus two plus three plus four plus five plus six. So, this is nothing but nn plus one by two. And what is n here? n is six. So, six plus one by two. That is three into seven. Three into seven. That's 21. So, our sixth term will be 21. So, option number C is correct for this question. So, hope everyone is clear on this. Let me increase some thickness for this. Okay. So, this is our first question. Okay. So, let's move to the next question. Question number two. So, what it is saying? The first three terms of a sequence are three comma three comma six. Okay. The second term after the second is the sum of the two terms preceding it. The eighth term of the sequences. Okay. So, one sequence is given whose first three terms are given. This three comma three comma six comma. This will proceed. And the question is asking to find the eighth term. Okay. So, the condition is given like each term after the second is the sum of the two terms preceding it. What does it mean? It means the nth term is equal to Tn minus two plus Tn minus one for n greater than two. Right. For n greater than two, this equation will hold. Okay. So, we have to find the eighth term. So, we can like in if this relation holds for n greater than two. So, we will write like if we put n equal to three, we get T3 equal to T1 plus T2. And what is T1? T1 we have three. T2 is also three. So, T3 will be six. If we put n equal to four, we will get T4 is equal to T2 plus T3. T2 and T2 is known to us T2 is three. So, three and T3 we derived in the last equation this six. So, T4 will be nine. If we put n equal to five, the fifth term we will be getting and this will be equal to T3 plus T4. T3 is six. T4 is nine. So, six plus nine. This will be 15. When we put n equal to six, we will get the sixth term as T4 plus T5. And we are having the value of T4 and T5 nine plus 15, that will be 24. When we put n equal to seven, T7 will come and it will be equal to T5 plus T6. This will be T5 plus T6 means 24 plus 15, that will be 39. And finally, when we put n equal to eight, we will get our eighth term which is asked in the question. It will be T6, sorry, it will be T6 plus T7. This will be 24 plus 39. So, nine, four, three, one, three, three, six. So, our eighth term is 63. This will be our answer. So, one condition was given in the question and we have done nothing. We have just followed that condition and we put the different values of n. And finally, we get the value of eighth term which was asked in the question. This is correct for this question. I hope everyone is getting. So, we can move to the next question. So, what was given here? If a n is equal to sine n pi by six, the value of sigma a n square n is going from one to six. So, this is given in terms of sine function that is sine n pi by six. Okay. So, the question is asking to find the value of this sigma n equal to one to six a n square, right? So, if we expand this sigma, we will get a one square plus a two square plus a three square plus a four square, a five square and a six square. n is going from one to six, okay? And what will be a one? Our a one will be a one will be sine. We will put one in place of this n. So, sine pi by six, sine pi by six is square. What will be a two? A two will be sine two pi by six and squared. What will be our a three? A three will be sine three pi by six squared. And similarly sine four pi by six squared. Then sine five pi by six squared and sine six pi by six squared. Okay. So, sine pi by six pi by six means 30 degree. So, sine pi by six will be one by two. So, this will be one by two square. Sine two pi by six is a sine 60 degree that is a root three by two squared. Then sine three pi by six sine 90 degree that will be one square. Sine four pi by six, sine one four into 30, sine 120 degree. This is that is nothing but cost 30 degree. That will be a root three by two squared. The sine pi by six will be one by two squared. And the sine six pi by six, sine pi is nothing but sine pi is zero. So, we will have here one by two square, one by two square. Like two times one by four plus two times. Sorry sir. What's going on? Sorry for the interruption. You think I have placed something. Okay. Never mind. I am not looking at this. So, two times one by four plus two times. It will be three by four plus one. So, this will be one by two plus three by two plus one. One by two plus three by two, two plus one, three. So, three will be answered to this question. Option number B is correct. Okay. So, we can move forward to the next question. Let's see what is the question. Okay. Here comes the question. What is it? If for a sequence a n, s n sum of n terms is given as two n square plus nine n. Okay. Where s n is the sum of n terms, the value of a 20. Okay. Some sequence is given, where sum of n terms is given to be two n square plus nine n. Okay. And we have to find the value of a 20. Means we have to find the 20th term. Okay. So, it's a simple, if we need to find the 20th term, what we have to do, we have to take the difference of sum of 20 terms and sum of 19 terms. Like, if s n is the sum of n terms to the sequence, how do we calculate the nth term? We calculate the nth term by taking the difference of sum of n terms and sum of n minus one terms. Okay. So, similar concept I have applied here also to find the 20th term and taking the difference of sum of 20 terms and sum of 19 terms. So, what will be s of 20? s of 20, you put n equal to 20 in this equation. We will get s 20. So, two into 20 square minus, sorry, plus nine into 20 and minus s 19. s 19 means we will put n equal to 19. So, two into 19 square plus nine into 19. So, take two common from these two terms. So, it will be two into 20 square minus 19 square and take nine common from these two terms. So, it will give nine 20 minus 19. Okay. So, it will be two into a square minus b square means 20 plus 19 and 20 minus 19 plus nine into one. So, this will be two into 39 into one plus nine. This will be two into 39, 78 plus nine. That will be nothing but 87. So, 87 is their option c. Option c is correct for this question. Okay. Let me, this question is done no anywhere. But let me tell you something else. The sum of n terms. As per question, the sum of n terms is given as two n square plus nine n. So, definitely this series a n will be in AP. This series, this, sorry, this sequence will be in AP. Why? Because the sum of n terms is of the form a n square. We know if the sum of n terms is of form a n square plus b n. This represents a AP and the nth term and the nth term of this series can be written as a into two n minus one plus b. Okay. So, these things I wish to like convey to all like this thing should be known by all the students who are like preparing for G and all competitive exams. So nth term will be given by a into two n minus one plus b. What is a here? A is two here because SN is two n square. The coefficient of n square is a basically so a is two here and the coefficient of the n is b. So, what is b? b is nine. b is nine. So, directly we can get the nth term as two into nth term means this 20th term. So, where is this? So, 20th term will be two into 40 minus one plus nine. So, this will be two into 39 plus nine. That will be nothing but two into 39, 78 plus nine, 87. So, anyway, you can go through either method either this or that you are going to find the same result. So, this question is done. Moving to the next question. So, this is the next question, question number five. I think some error is there in this question. Let me correct it. a1 equal to this, a2 is this. an equal to two times an minus one plus five for n greater than two. This will be two for n greater than two. In question it was given as n greater than one, but n should be greater than two. So, what is the question? If a1 is equal to two, a2 is equal to three plus a1 and nth term is given as two an minus one plus five for n greater than two. The value of sigma a r, r from r is going from two to five. Okay. So, what is given? Let me write first. So, a1 is given to be, a1 is given to be two. a2 is given as three plus a1 and an is given as two times an minus one plus five for n greater than two. Okay. And what is asked basically? We have to find this sigma a r where r is going from two to five. So, if we expand it, we need to find the summation of a2 plus a3 plus a4 plus a5. Where a1, where a2 is known to us, what is a2? a2 is three plus a1, a1 is known to, so a2 will be basically three plus two. That is nothing but five. Let me find the value of a3. So, putting n equal to three, we can find a3. So, what will be a3? a3 will be equal to two into a2 plus five. And we know the value of a2, a2 is five. So, it will be ten plus five is equal to fifteen. And what will be a4? a4 will be two times a3 plus five. a3 is fifteen. So, two times of a3 will be thirty plus five. This will be thirty-five. And what will be a5? a5 will be two times a4 plus five. Two times a4 will be two into thirty-five. That will be seventy plus five is equal to seventy-five. So, we are done. Just we have to add the values of a2, a3, a4 and a5. So, a2 is five. a3 is fifteen. a4 is thirty-five. a5 is seventy-five. So, five plus fifteen. This will be twenty plus seventy-thirty-hundred. Hundred ten plus twenty. That will be one thirty. So, this will be answered to these questions. So, option number A is correct for this question. So, I think we are done with this exercise. Exercise number one. In the subsequent videos, in the coming videos, I will be again, I will be providing solutions for the next exercise like exercise two and exercise three. So, in similar fashion, this video series will continue. And I suggest to utilize this video series to the maximum so that you can gain confidence in preparing and in solving this. So, that's all for today. And soon we will meet again. So, thank you. Thank you, everyone. Okay.