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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