 Hi and welcome to the session. My name is Shashi and I am going to help you to solve the following question. Question is how many 3 digit numbers are divisible by 7? Let us start with the solution now. You know smallest 3 digit number divisible by 7 is 105 and the largest 3 digit number divisible by 7 is 994. So we have to determine the number of terms in the sequence 105, 112, 119, till 994. Now in this sequence the first number is the smallest 3 digit number which is divisible by 7 and the last number of the sequence is the largest 3 digit number which is divisible by 7. So clearly this sequence is a AP sequence as the common difference between the consecutive terms is equal to 7. So we can write it is an AP with first term A is equal to 105 and the common difference D is equal to 7. Now we know the nth term of an AP is given by An is equal to A plus n minus 1 multiplied by D where A is the first term of the AP and D is the common difference. Now let N be the number of terms in the AP. Then nth term is equal to that is An is equal to the last term of the AP that is 994 and we know An is given by A plus n minus 1 multiplied by D. Now we will substitute for An 994. We will substitute for A 105, N is unknown and D is equal to 7. So we will substitute for D 7. Now this implies 994 minus 105 is equal to 7 multiplied by N minus 1. Now we get 889 is equal to 7 multiplied by N minus 1. Now this implies 889 upon 7 is equal to N minus 1 dividing both sides by 7. We get 889 upon 7 is equal to N minus 1. Now this implies 127 is equal to N minus 1. We know 7 multiplied by 127 is equal to 889. Now we can write N minus 1 is equal to 127. This implies N is equal to 127 plus 1 or N is equal to 128. So we assume that N was the number of terms in the given AP. So total terms in the given AP is equal to 128. Hence there are 128 3 digit numbers divisible by 7. So our required answer is 128. This completes the session. Have a nice day. Goodbye.