Find the sum of all three digit numbers which are multiples of 9.

First three-digit number which is a multiple of 9 = 108

Second three-digit number which is a multiple of 9 = 117

Third three-digit number which is a multiple of 9 = 126

…

n^th three digit number which is a multiple of 9 = 999

So, the arithmetic sequence is 108, 117, 126, …, 999

Common difference = a = 117 − 108 = 9

First term of the arithmetic sequence = a + b = 108

⇒ 9 + b = 108

⇒ b = 108 − 9 = 99

Therefore,

⇒ 999 = 9n + 99

⇒ 999 − 99 = 9n

⇒ 900 = 9n

⇒ n = 100

Therefore, there are 100 three-digit numbers, which are the multiples of 9.

We know that the sum of a specified number of the consecutive terms of an arithmetic sequence is half the product of the number of the terms with the sum of the first and the last term.

∴ Sum of its first 100 terms = … (1)

Here,

Putting the values in equation (1):

Sum of first 100 terms =