Question

Find the sum of all 3 digit natural numbers which are multiples of 9.​

Answer 1

Step-by-step explanation:

The least 3 digit number divisible by 9 is 108

The greatest 3 digit number divisible by 9 in999

a=108,l=999 in order to find n we apply tn formula

Step-by-step explanation:

tn=a+(n-1)d

999=108+(n-1)9

999-108=n-1×9

891/9=n-1

99=n-1

100=n

Now let's find the sum

sn=n/2(a+l)

=100/2(108+999)

=50(1107)

=55350

Hope it helps

Answer 2

Answer:

55350

Step-by-step explanation:

the least three digit number divisible by 9 is 108 and the greatest three digit number is 999.

a=108 ; l=999

tn=a+(n-1)d

999=108+(n-1)9

999-108=n-1x9

891/9=n-1

99=n-1

99+1=n

n=100

Now the sum,

sn=n/2(a+l)

100/2(108+999)

50(1107)

55350