find the number of all 3 digit natural nos. which are divisible by 9
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101,102,103,.... 999 are 3 digit numbers
108, 117,126, ... ,999 are all 3 digit numbers which are divisible by 9[ add 9 for each number u can get next number] 108,117,126 ...,999 are in AP first term = a =108, common difference = d = a2 -a1 =117 -108 =9, last term = l =999, let the number of terms = n, last term = 999, a+(n-1)d =999, 108+(n-1) 9= 999, (n-1)9 = 999-108, (n-1)9 =891, n-1 = 891/9, n-1 = 99, n= 99+1, n= 100
108, 117,126, ... ,999 are all 3 digit numbers which are divisible by 9[ add 9 for each number u can get next number] 108,117,126 ...,999 are in AP first term = a =108, common difference = d = a2 -a1 =117 -108 =9, last term = l =999, let the number of terms = n, last term = 999, a+(n-1)d =999, 108+(n-1) 9= 999, (n-1)9 = 999-108, (n-1)9 =891, n-1 = 891/9, n-1 = 99, n= 99+1, n= 100
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