 Hello and welcome to the session. Let us discuss the following problem today. In an AP given A is equal to 2, D is equal to 8, SN is equal to 90, find N an AM. Now let us write the solution. We know SN is equal to N by 2 multiplied by 2A plus N minus 1 multiplied by D. Now substituting the values SN is given to us as 90 which is equal to N by 2 multiplied by 2 multiplied by A is equal to 2 plus N minus 1 multiplied by D which is equal to 8. So we can see here only 1 N is unknown so we will find N. Taking these two here so we get 1AT is equal to N multiplied by 4 plus N minus 1 multiplied by 8. Which implies 1AT is equal to N multiplied by 4 plus 8N minus 8 which implies 1AT is equal to N multiplied by 8N 4 minus 8 is minus 4. Which implies 1AT is equal to 8N square minus 4N which implies 8N square minus 4N minus 1AT is equal to 0. Now taking 2 common we get 4N square minus 2N minus 90 is equal to 0. Now again taking 2 common we get 2N square minus N minus 45 is equal to 0. Now solving this on splitting the middle term we get 2N square minus 10N plus 9N minus 45 is equal to 0. Now from these two terms taking 2N common so we get N minus 5 plus 9 taking 9 common so we get 9 minus 5 is equal to 0. Which implies N minus 5 multiplied by 2N plus 9 is equal to 0. Which implies N minus 5 is equal to 0 or 2N plus 9 is equal to 0. Which implies N is equal to 5 and N is equal to minus 9 by 2. But N is equal to minus 9 by 2 is not possible because number of terms cannot be negative and infractions. So therefore N is equal to 5. Now we know AN is equal to A plus N minus 1D. Now we have to find the value of AN so substituting the values 2 plus 5 minus 1 multiplied by 8. Which is equal to 2 plus 4 into 8. Which is equal to 2 plus 32 which is equal to 34. Therefore AN is equal to 34. Hence N is equal to 5 and AN is equal to 34 is our required answer. I hope you understood the problem bye and have a nice day.