Question

Find the sum of all the three digit natural numbers which are multiple of 9

Answer 1

Answer:

First three digit number divisible by 9 is 108.

⇒ a = 108

⇒ d = 9

⇒ n = ?

⇒ L = 999

Applying the formula: L = a + ( n - 1 ) d , we get,

⇒ 999 = 108 + ( n - 1 ) 9

⇒ 999 - 108 = ( n - 1 ) 9

⇒ 891 = ( n - 1 ) 9

⇒ 891 / 9 = ( n - 1 )

⇒ 99 = ( n - 1 )

⇒ 99 + 1 = n

⇒ 100 = n

Hence the number of terms is 100.

Applying Sum formula we get,

$\implies S_{n} = \dfrac{n}{2}\: [ \: a + L \: ]\\\\\implies S_{100} = \dfrac{100}{2} \: [ \: 108 + 999 \: ] \\\\\implies S_{100} = 50 \: [ 1107 \: ]\\\\\implies S_{100} = 55350$

Hence the sum of all three digit natural numbers which are a multiple of 9 is 55,350

Answer 2

Answer:

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