 Hello and welcome to the session. In this session we discussed the following question which says, write the given set in roster form. We have the set A is equal to X such that X is equal to 2Y minus 3 where Y belongs to N and Y is greater than equal to 1 and less than 5. Then in the next part we have, write the given set in set builder form. We have the set B is equal to the set containing the elements 39, 27, 81 and 243. Let's start with the solution now. In the first part we are given a set A equal to X such that X is equal to 2Y minus 3 where Y belongs to N that is the set of natural numbers and Y is greater than equal to 1 and less than 5. We need to write the set in the roster form. In the roster form we write the elements of the set within the braces. So first of all let's find out the elements of the set A. We are given Y belongs to the set of natural numbers and Y is greater than equal to 1 and less than 5. Therefore Y takes the values 1, 2, 3 and 4. Also we have X is equal to 2Y minus 3. Now since we have the set A is equal to X, given the conditions for X and Y. So this means that we will find the values of X to find out the elements of the set A. Now for Y equal to 1 we have X is equal to 2 multiplied by 1 minus 3 that is we substitute Y equal to 1 in this. So this means we get 2 minus 3 that is minus 1. Now for Y equal to 2 we have X equal to 2 multiplied by 2 minus 3 that is 4 minus 3 which is equal to 1. Let's now find out the value of X for Y equal to 3. We get X equal to 2 multiplied by 3 minus 3 that is 6 minus 3 which is 3. Then again for Y equal to 4 we have X equal to 2 multiplied by 4 minus 3 that is 8 minus 3 which is 5. Thus we have putting Y equal to 1, 2, 3, 4 we obtain X equal to minus 1, 1, 3, 5. So we were supposed to find the values of X to get the elements of A. So we have got the values of X for the values of Y. Thus we can say in the roaster form the set A is equal to the set containing the elements minus 1, 1, 3, 5. So we have got the set A in the roaster form. So this is the answer for the first part of the question. Now next we have set B containing the elements 3, 9, 27, 81, 243. We need to write this set in set builder form. In the set builder form the set is denoted by using words, formulas or properties. Now to write this given set in the set builder form we need to observe each element of this set. First we have the element 3. This can be written as 3 to the power of 1. Then we have 9 which could be written as 3 to the power of 2. Then we have 27 which is the third element of this set. This could be written as 3 to the power of 3. Then we have the element 81 which is 3 to the power of 4. Then we have the element 243 which is 3 to the power of 5. So we observe that each element of the set B is written as some power of 3. So this means in set builder form we have B is equal to x such that x is equal to 3 to the power of n plus 1 where n belongs to w that is the set of whole numbers and n is less than 5. So this is our answer for the second part of the question. With this we complete the session. Hope you have understood the solution of this question.