ARITHMATIC PROGRESSION

Asked by chahat | 11th Mar, 2009, 12:36: AM

Expert Answer:

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. 108+117+126+...+999 let 999 be the nth term  first term =a=108, d=9 so ,999=a+(n-1)d so,  999=108+9(n-1) 891=9(n-1) n-1=99 so n=100 required sum= n/2[a+l} where l is the last term of the A.P. =100/2[108+999] 50*[1107] multiply these two numbers and you will have the answer.

Aswered by  | 11th Mar, 2009, 09:30: AM

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