Question

calculate sum of all three digit numbers which are multipile of 9

Answer 1

Step-by-step explanation:

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Answer 2

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,

\begin{gathered}\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\end{gathered}

⟹S

n

=

2

n

[a+L]

⟹S

100

=

2

100

[108+999]

⟹S

100

=50[1107]

⟹S

100

=55350