 Hello and welcome to the session. I am Arsha and I am going to help you with the following question which says, find the sum of all two digit numbers which when divided by four yields one as remainder. And with the solution, two digit numbers which when divided by four and yielding remainder one, the first number is 13, the next number is 17 and we have 21 and the last number is 97. So this is an AP series whose first term, it is denoted by small a is 13 and the common difference, it is denoted by small a is equal to 4 and the last term which is an is equal to 97. Now an is a plus n minus 1 into d where n is the number of terms and the sequence. Now let us substitute the values of a and d to get the value of n. So a is 13 plus n minus 1 into 4 is equal to 97 or we have n minus 1 into 4 is equal to 97 minus 13 gives 84 or n minus 1 is equal to 84 upon 4 which is equal to 21 or we have n is equal to 21 plus 1 that is 22. And now we have to find the sum of these 22 terms of the sequence. So now s n is n upon 2 into 2 a plus n minus 1 into d. So to find the sum of 22 terms we have s 22 is equal to 22 upon 2 into 2 into a is first term which is 13 plus 22 minus 1 into the common difference which is 4 which is further equal to 11 into 2 into 13 is 26 plus 21 into 4 which is equal to 11 into 26 plus 84 or 11 into 110 which is further equal to 1210 therefore the required sum is 1210. So this completes the session take care and have a good day.