 Hello and welcome to the session. Let us understand the following question today. Find the sum of the first 40 positive integers divisible by 6. Now let us write the solution. The first positive integer divisible by 6 is 6. The second positive integer divisible by 6 is equal to 12. Similarly, the third positive integer divisible by 6 is equal to 18 and so on we can find the positive integer divisible by 6. Thus the AP formed is 6, 12, 18, 24. Here we see that is equal to 6. D is equal to 12 minus 6 which is equal to 6. N is equal to 40 since we have to find the sum of first 40 positive integers divisible by 6. So, N is equal to N by 2 multiplied by 2A plus N minus 1D. Now we have to find S40 which is equal to 40 by 2 multiplied by 2 into 6 plus 40 minus 1 multiplied by 6. Now we see that this gets cancelled by 20 which is equal to 20 multiplied by 12 plus 39 into 6 which is equal to 20 multiplied by 12 plus 234 which is equal to 20 multiplied by 246 which is equal to 4920. Therefore, divisible by 6 is equal to 4 now a required answer. I hope you understood the question. Bye and have a nice day.