Answer :

Listing all 3-digit natural numbers which are multiples of 9 are –

108, 117, 126, … , 999

We have to find 108 + 117 + 126 + … + 999

This is clearly an AP.

Here, a = 108, d = 9 and a_{n} = 999 …(i)

Lets find n.

a + (n – 1)d = a_{n} …(ii)

Substituting equation (i) in (ii),

108 + (n – 1)×9 = 999

⇒ 9n – 9 = 999 – 108 = 891

⇒ 9n = 891 + 9 = 900

⇒ n = 900/9 = 100

There are 100 (=n) numbers in the series.

Now sum of this AP is given by

S_{100} = (n/2)[a + a_{n}]

⇒ S_{100} = (100/2)[108 + 999]

⇒ S_{100} = 50 × 1107 = 55350

Thus, the sum of all three-digit natural numbers, which are multiples of 9 is 55350.

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