Sum of all 3 digit numbers which are the multiple 9.

Question

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

Answer 1

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, 

x_n = 9n + 99, where n is a natural nimber.

⇒ 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 = 1/2 x 100 x (x₁₀₀ + x₁) … (1)

Here, 

x₁ = 108 and x₁₀₀ = 999

Putting the values in equation (1):

Sum of first 100 terms = 1/2 x 100 x (999 + 108)

= 50 x 1107

= 55350