 Hello and welcome to the session. The given question says, find the sum of all the natural numbers less than 100 which are divisible by 6. Let's start with the solution and we have to find the sum of all the natural numbers less than 100 which are divisible by 6. So the natural numbers divisible by 6 less than 100 but the first number is 6 itself since it's the smallest natural number which is divisible by 6 then we have 12, 18, 24, 30, 36 and so on up to the last number which is divisible by 6 and less than 100 is 96. Before that we have 90 and so on. Now 6 is the first term and 96 is the nth term. Now we know that An is given by A plus n minus 1 into D where A is equal to A1 which is the first term of the AP series. This is the AP series. As we can see the common difference between each of these terms is 6 which is the common difference and here it is equal to 6 and n is the number of terms in this AP series. Now here we have Anx term is equal to 6 plus n minus 1 into 6 where 6 is the common difference and this 6 is the first term. Suppose we are taking the last term as the nth term to get the value of n then we have 96 is equal to 6 plus n minus 1 into 6. We are trying to find the number of terms here. So this implies that 90 is equal to n minus 1 into 6 or n minus 1 is equal to 90 divided by 6 which is equal to 15 or n is equal to 16. So here the number of terms of this AP series is equal to 16. Denoting it by small n it is equal to 16. Now we have to find the sum of these 16 terms. So we have as we know the sum of n terms is given by n divided by 2 into 2A plus n minus 1 into D and here we have to find the sum of first 16 terms. So we have 16 divided by 2 into 2 times of 6 plus 16 minus 1 into 6 which is further equal to 8 into 12 plus 15 into 6 and this is further equal to 8 into 12 plus 15 6 is 90. This is equal to 8 into 102 which is further equal to 816. Hence the answer is the sum of all the natural numbers less than 100 which are divisible by 6 is 816. So this completes this asian by intake