Question

The sum of all 3-digit natural numbers which are divisible by 9 is
54243
54351
55350
55242

Answer 1

$\large\underline{\sf{Solution-}}$

The 3 digit natural numbers which are divisible by 9 are

$\rm :\longmapsto\:108, \: 117, \: 126, \: - - - - - , \: 999$

So,

It implies, its an AP series as common difference is same.

Thus,

First term of AP, a = 108

Common difference, d = 9

nᵗʰ term of series = 999

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Tʜᴜs,

On substituting the values, we get

$\rm :\longmapsto\:999 = 108 + (n - 1)9$

$\rm :\longmapsto\:999 - 108 = (n - 1)9$

$\rm :\longmapsto\:891= (n - 1)9$

$\rm :\longmapsto\:99= n - 1$

$\bf\implies \:n = 100$

Now,

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(\:a\:+a_n \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.

• a is the first term of the sequence.

• n is the no. of terms.

• aₙ is the nᵗʰ term.

Tʜᴜs,

$\rm :\longmapsto\:S_n = \dfrac{100}{2} (108 + 999)$

$\rm :\longmapsto\:S_n = 50 \times 1107$

$\bf :\longmapsto\:S_n = 55350$

Additional Information :-

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference