Asked by chahat | 11th Mar, 2009, 12:36: AM
the three digit numbers which are divisible by 9 start with 108 and end with 999.
These numbers obviously are such that there is a gap of 9 between the consecutive numbers.
So, these numbers form an A.P.
So, basically we have to find the sum of the A.P.
let 999 be the nth term
first term =a=108, d=9
n/2[a+l} where l is the last term of the A.P.
multiply these two numbers and you will have the answer.
Answered by | 11th Mar, 2009, 09:30: AM
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