 Hello and welcome to the session. I am Ashima and I am there to help you with the following problem. Find the sum of the following AP. We have 1 by 15 comma 1 by 12 comma 1 by 10 so on till 11 terms. Now let us write the solution. Given to us AP as 1 by 15 1 by 12 1 by 10 so on to 11 terms. Here A is equal to 1 by 15, D is equal to common difference so 1 by 12 minus 1 by 15 which is equal to now taking the same we get 60, 12 5 is the 60 so 5 minus 15 4 is also 1, 4 is also 4 so which is equal to 1 by 60. Now N is equal to 11. Now applying the formula of sum of AP which is equal to N by 2 multiplied by 2A plus N minus 1 D. Now here N is equal to 11 so 11 by 2 multiplied by 2 into 1 by 15 plus 11 minus 1 multiplied by 1 by 60. Now which is equal to 11 by 2 multiplied by 2 by 15 plus 10 multiplied by 1 by 60 which is equal to 11 by 2 multiplied by 2 by 15 plus this gets cancelled so we have 1 by 6. Now taking LCM we get which is equal to 11 by 2 so 30 is the LCM here 15 to the 30 so 2 2s of 4 plus 6 5s of 30 so 1 5 is a 5 which is equal to 11 by 2 multiplied by 9 by 30 so here this gets cancelled by 3 and this by 10 so we get 33 by 20 therefore required sum is 33 by 20. I hope you understood this problem bye and have a nice day.