Question

Find the sum of all natural numbers which are multiples of $7$ or $3$ or both and lie between $200$ and $500$.

Answer 1

Proceeding as in Q, we have
$S=S_3+S_7-S_{21}$
$S_3=$ sum of all those number between $200$ and $500$ which are divisible by $3$
$=201+204+..+498$
$498=201+(n-1)3$ by $l=a+(n-1)d$
$\therefore n=100$
$\therefore S_3=\frac{100}{2}[201+498]=50\times 699=34950$
$S_7=203+210+....+497$
$497=203+(n-1)7$ or $\frac{294}{7}=n=1$ $\Rightarrow n=43$
$S_7=\frac{43}{2}[201+497]=15050$
$S_{21}=210+231+...+483$
$483=210+(n-1)21$
$\therefore \frac{273}{21}=n-1$
$\Rightarrow n=14$
$\therefore S_{21}=\frac{14}{2}[210+483]=4851$
$\therefore S=S_3+S_7-S_{21}=34950+15050-4851$
$=50000-4851=45149$.