 Hello and welcome to the session. In this session we will discuss a question which says that if the same as the second and eighth terms of an AP is 22 and their product is 85, find the AP. Now before starting the solution of this question we should know some results. And first is the nth term of an AP is given by a n is equal to a plus n minus 1 divided by into t where a is the first term of an AP and d is the common difference. And secondly an AP can be written as a plus d, a plus 2d, a plus 3d and so on where a is the first term and d is the common difference of the AP. Now these results will work out as a key idea for solving out this question. And now we will start with the solution. Now first of all we will find the second and eighth terms of an AP. Now using this result which is given in the key idea the second term of an AP that is a2 is equal to plus 2 minus 1 the whole into t where a is the first term and d is the common difference. So this implies a2 is equal to a plus d. Now an AP that is a8 is equal to a plus a is minus 1 the whole into t which implies a8 is equal to a plus 7d. Now it is given that the sum of the second and eighth terms of an AP is 22. So given which is the second term plus a8 which is the eighth term is equal to 22. Now this is the value of a8 and this is the value of a2. So given these values here this implies a plus d plus a plus 7d is equal to 22 which further implies a t is equal to 22. Now dividing throughout by 2 this implies a plus 4d is equal to 11 which implies a is equal to 11 minus 4d. And let us name it as 1. Also it is given in the question that the product of the second and eighth terms is 85. So given the second and eighth term is equal to 85. Now continuing the values of a2 and a8 here it will be a plus d the whole into a plus 7d the whole is equal to 85. Now from then the value of a is this. So putting this value of a here this implies 11 minus 4d plus d the whole into 11 minus 4d plus 7d the whole is equal to 85. Which further implies 11 minus 3d the whole into 11 plus 3d the whole is equal to 85. Now applying the value of a minus b the whole into a plus b the whole this implies 11 square minus 3d square is equal to 85. Which further implies 129 minus 9d square is equal to 85. Which further implies 9d square is equal to 121 minus 35. Which implies 9d square is equal to 36 which further implies d square is equal to 36 by 9 which is equal to 4 which further implies d is equal to plus minus 2. Now this is the equation number 1. Now for d is equal to plus 2 here is equal to 11 minus 4d will be equal to 11 minus 4 into 2 which is equal to 11 minus 8 which is equal to 3. And for d is equal to minus 2 here will be equal to 11 minus 4d which is equal to 11 minus 4 of minus 2 which is equal to 11 plus 8 which is equal to 19. Now using this result which is given in the key idea here ap is a plus d a plus 2d and so on. And putting the values of a and d here a in this case is 3 and d is 2. So this will be 3 into 2 and so on which is 3 5 3 plus 4 which is 7. Now here d is equal to minus 2 and a is equal to 19. So here ap is a a plus d a plus 2d and so on. Now putting the values of a and d here it will be 19 19 plus of minus 2 19 plus 2 of minus 2 and so on which will be 19 and 19 minus 2 will be 17 and here 19 minus 4 will be equal to 15 and so on. So these are the required ap. So this is the solution of the given question and that's all for this session. Hope you all have enjoyed this session.