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Question -

Find the number of ail three digit natural numbers which are divisible by 9. [CBSE 2013]



Answer -

First 3-digit numberwhich is divisible by 9 = 108
and last 3-digit number = 999
d= 9
a + (n – 1) d = 999
=> 108 + (n – 1) x 9 = 999
=> (n – 1) d = 999 – 108
=> (n – 1) x 9 = 891
=> n – 1 = 99
=> n = 99 + 1 = 100
Number of terms = 100

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