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.

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

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