find the sum of all, 3-digit natural nos. which
are divisible.9
The
Answers
Answer:
here is your ans
Explanation:
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
Hence the sum of all three digit natural numbers which are a multiple of 9 is 55,350
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Answer:
55350
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