Request a call back

ARITHMATIC PROGRESSION
Asked by chahat | 11 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.

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 | 11 Mar, 2009, 09:30: AM

## Application Videos

CBSE 10 - Maths
Asked by devanshsinghbais111 | 11 Oct, 2022, 01:57: PM
CBSE 10 - Maths
Asked by Subuhiansari67 | 28 Sep, 2022, 10:11: PM
CBSE 10 - Maths
Asked by Bavashreelogasundaram | 22 Sep, 2022, 08:26: PM
CBSE 10 - Maths
Asked by srinew414 | 03 Sep, 2022, 12:16: PM
CBSE 10 - Maths
Asked by sharmaraushan079 | 19 Aug, 2022, 04:04: PM
CBSE 10 - Maths
Asked by Sunita | 08 Aug, 2022, 01:06: PM
CBSE 10 - Maths
Asked by choithanimahak | 05 Aug, 2022, 09:06: PM