Q8 of 46 Page 1

Find the sum of all three digit natural numbers, which are multiples of 11.

We need to find 110 + 121 + … + 990.

Let


a = first number of the series,


d = the constant difference between each number,


n = total numbers in the series, and


an = the last number of the series


Here, a = 110, d = 121 – 110 = 11 and an = 990


First, we need to find n.


an = a + (n – 1)d


⇒ 990 = 110 + (n – 1)×11


⇒ 990 = 11 (10 + n – 1)


⇒ 90 = 9 + n


⇒ n = 90 – 9 = 81


The series contains 81 numbers, i.e., n = 81.


Now, we can find the sum of these 81 numbers, which is given by


Sum =


⇒ Sum = = = 81×550 = 44550


Hence, sum of all 3-digit natural numbers which are multiples of 11 is 44550.


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