Question

Find the sum of all three digit natural number divisible by $3$.

Answer 1

Three digit numbers divisible by $3$ are $102,105,.....,999$

It is nothing but a series of AP whose common difference is $3$ and first term is $102$

$\Rightarrow$ A.P with $a=102$, $d=3$, $a_n=999$

Apply the formula for $n^{th}$ term of an AP

$\Rightarrow 102+(n-1)3=999$

$\Rightarrow 3(n-1)=897$

$\Rightarrow n-1=299$

$\Rightarrow n=300$

Now, apply the formula for sum of first $n$ terms of an AP

$\Rightarrow S_n=\frac{n}{2}(a+a_n)$

$=\frac{300}{2}(102+999)$

$=1101\times 150$

$=165150$.