Answer :

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

a_{n} = the last number of the series

Here, a = 110, d = 121 – 110 = 11 and a_{n} = 990

First, we need to find n.

a_{n} = 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.

Rate this question :

How useful is this solution?

We strive to provide quality solutions. Please rate us to serve you better.

Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Expertsview all courses

Dedicated counsellor for each student

24X7 Doubt Resolution

Daily Report Card

Detailed Performance Evaluation