Q. 145.0( 3 Votes )

# Find the sum of a

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 an = 999 …(i)

Lets find n.

a + (n – 1)d = an …(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

S100 = (n/2)[a + an]

S100 = (100/2)[108 + 999]

S100 = 50 × 1107 = 55350

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

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