Question

Find the sum of all $3$ digit natural multiples of $6$.

Answer 1

Sum of all 3 digit natural multiple

of 6 is $82,350$

3 digit natural no. multiple of 6 are

$102,108,114,\mathrm{120....}996$

$a=102$

$d=6$

Let the last term be nth term

$996=102+(n-1)6$

$996-102=(n-1)6$

$(n-1)6=894$

$n-1=\frac{894}{6}$

$n-1=149$

$n=149+1$

$n=150$

Sum of all 3 digit no. multiple of $6=\frac{n}{2}(a+1)$

$=\frac{150}{2}(102+996)$

$=150\left(549\right)$

$=82,350$