Q. 85.0( 6 Votes )

# Find the sum of a

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

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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