An important and exam oriented question from arithmetic progression chapter as it was already asked in previous year paper of 2012 in which we have been asked to find the number of 3-digit number divisible by 9.
RS Aggarwal, class 10, arithmetic progression, chapter 5A, question no 44
The three digits numbers which are divisible by 9 are 108,117,126,…,999
Then, first term a =108 Common difference =9
Last term =999 We know that, l=an=a+(n−1)d
999=108+(n−1)9
999−108=9n−9
891+9=9n
900=9n
n=900/9
n=100
Therefore, there are 100 three digits numbers.