 Hello and welcome to session. I am Bhumika and we'll discuss with you first part of question 14 from exercise 5.1 which is at page 90 of your book. Now the question says find the sum of the first 11 terms of the arithmetic progression 2, 6, 10 and so on. Now here we'll be using the sum of the first n terms formula to do these questions. Now we know if a is the first term of a p and d is the common difference of a p then sum of first n terms Sn the notation is Sn. This is a standard notation Sn is equal to n by 2 2a plus n minus 1 into d this is a formula to find the sum of first n terms where a is the first term and d is the conference and we will be using this as our key idea in solving this. So let's begin our solution. What we have a series. We have a sequence arithmetic progression We have 2, 6, 10 a is 2 and d is 6 minus 2 which is 4 because d is the difference between two conjugative term. Therefore it's 6 minus 2 which is 4, 10 minus 6 which is also 4 You can take any difference. So d is 4, a is 2. Now what the question says is we have to find the first 11 terms of the a p. You have to find the sum of the first 11 terms. Here n is 11 we know that Sn is equal to n by 2 2a plus n minus 1 into d Therefore S11 is equal to 11 by 2 into 2a plus 10d Substitute the values 2 into 2 plus 10 into 4. How you got plug-in values After solving we got this is 4. This is 40 which is equal to 11 by 2 in 44 which is equals to 11 into 22 Hence the answer is 242 Therefore, what is the answer? Hence the sum of first 11 terms Is 242 This is the answer. So what it it's very very simple question Direct application of formula Sn is equal to n by 2 into 2a plus minus 1 into d just remember this formula by heart by heart and Any question of this type will be done through this formula and just revise it once again and keep remember this formula So that's all from this session. Have a nice time. Take care