Question

Find the sum of all the multiples of $3$ digits numbers which are divisible by $\left(a\right)6,\left(b\right)9$

Answer 1

$\left(a\right)$ $3$ digit numbers divisible by $6$:

$\Rightarrow 102,\mathrm{108...996}$

Here $a=102,d=6$

$\Rightarrow t_n=a+(n-1)d$

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

$\Rightarrow (n-1)6=894$

$\Rightarrow (n-1)6=894$

$\Rightarrow n=150$

$\Rightarrow S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

$\Rightarrow S_n=\frac{150}{2}(2\times 102+149\times 6)$

$\Rightarrow S_n=82350$

$\left(b\right)$ $3$ digit numbers divisible by $9$:

$\Rightarrow 108,\mathrm{117...999}$

Here $a=108,d=9$

$\Rightarrow t_n=a+(n-1)d$

$\Rightarrow 999=108+(n-1)9$

$\Rightarrow (n-1)9=891$

$\Rightarrow n=100$

$\Rightarrow S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

$\Rightarrow S_n=\frac{100}{2}(2\times 108+99\times 9)$

$\Rightarrow S_n=55350$