 Hi and welcome to the session. I am Neha and I am going to help you with the following question. The question says whether the following list of numbers form an AP. If they form an AP, find the common difference D and write 3 more terms. 2, 5 by 2, 3, 7 by 2 and so on. Before proceeding for the solution, recall that a sequence A1, A2, A3 and so on form an AP with the difference of a term and its preceding term is always constant. And this constant is called the common difference of the AP. This is the key idea for this question. Now let's see its solution. We are given the sequence 2, 5 by 2, 3, 7 by 2 and so on. So that means here this is A1, A2, A3 and A4. Now we need to find out whether the sequence form an AP or not. So according to the key idea, we will find out the difference of the term and its preceding term. So first of all A2 minus A1 will be equal to 5 upon 2 minus 2 which is equal to 1 upon 2. Next A3 minus A2 is equal to 3 minus 5 by 2 which is also equal to 1 by 2. Now A4 minus A3 is equal to 7 by 2 minus 3 which is also equal to 1 by 2. Now as we can see that the difference of the term and its preceding term is always the same that is 1 by 2. So we have A k plus 1 minus A k is always constant. Therefore the sequence 3, 7 by 2 and so on form an AP. And the common difference D is equal to 1 by 2. Now we need to find 3 more terms. So for this sequence the fifth term will be A4 that is 7 by 2 plus common difference that is 1 by 2. So here we have A5 is equal to 7 by 2 plus 1 by 2 which is equal to 8 by 2 that is 6 term. A6 will be equal to A5 that is 4 plus 1 by 2 which will be equal to 9 by 7 will be equal to A6 that is 9 by 2 plus 1 by 2 which will be equal to 10 by 2 that is it. We have found the next three terms and thus the answer to this question is yes the sequence forms an AP. The common difference D is equal to 1 by 2 and the three more terms are 4, 9 by 2. Thus we finish this session. Hope you must have understood the question. Goodbye, take care and have a nice day.