 Hello and welcome to the session. I am Asha and I am going to help you with the following question which says, find the sum of all the natural numbers lying between 100 and 1000 which are multiples of 5. So let us begin with the solution and we have to find the sum of all the multiples of 5 which are between 100 and 1000. So the first number is 105, next is 110 plus 115 plus so on up to 995. So here the first term is equal to 105, common difference is equal to 110 minus 105 is 5 and we have to find n first and then we will find the sum. The last term whose formula is a plus n minus 1 into d where d is the common difference. So the last term of this sequence is 995 which is equal to a is 105 plus n minus 1 into d is common difference is 5. So here 995 minus 105 is equal to n minus 1 into 5 is equal to on subtracting 105 from 995 we get 890 is equal to n minus 1 into 5 or n minus 1 is equal to 890 upon 5 or we have n minus 1 is equal to 5 1s of 5 3 5 7s of 35 4 5 8s of 40 so n is equal to 179. Now we have to find 3 equal to n 2 minus 1 into d this is the sum terms which are 179 in number so this is equal to 179 upon 2 2 into 105 plus 179 minus 1 into d which is 5 so this is equal to 179 upon 2 105 into 2 is 210 plus 178 into 5 which is equal to 179 upon 2 into 210 plus 890 which is equal to 179 upon 2 into 1100 now to find the 10 again to find the 10 to 0 this is equal to 179 into 550 which is equal to 98450 so the sum of all the natural numbers 9 between 100 and 1000 which are multiples of 5 is 98450 so this complete calculation take care and have a good day.