$\mathrm{Find} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{natural} \mathrm{numbers} \mathrm{between} 250 \mathrm{and} 1000 \mathrm{which} \mathrm{are} \mathrm{exactly} \mathrm{divisible} \mathrm{by} 3.$

$$\begin{gathered} \mathrm{The} \mathrm{numbers} \mathrm{between} 250 \mathrm{and} 1000 \mathrm{which} \mathrm{are} \mathrm{divisible} \mathrm{by} 3 \\ \mathrm{are} 252, 255, 258, \dots ., 999. \end{gathered}$$

$$\begin{gathered} \mathrm{This} \mathrm{is} \mathrm{an} \mathrm{A}.\mathrm{P}. \mathrm{with} \mathrm{first} \mathrm{term} \mathrm{a} = 252, \\ \mathrm{common} \mathrm{difference} = 3 \mathrm{and} \mathrm{last} \mathrm{term} = 999. \end{gathered}$$

$$\begin{gathered} \mathrm{Let} \mathrm{there} \mathrm{be} \mathrm{n} \mathrm{terms} \mathrm{in} \mathrm{this} \mathrm{A}.\mathrm{P}. \\ \mathrm{Then}, = a_n = 999 = \mathrm{a} + (\mathrm{n} – 1)\mathrm{d} = 999 \end{gathered}$$

$$\begin{gathered} = 252 + (\mathrm{n} – 1) \times 3 = 999 \\ \Rightarrow \mathrm{n} = 250 \end{gathered}$$

$\therefore \mathrm{Required} \mathrm{sum} = {\mathrm{S}}_{\mathrm{n}} = \frac{n}{2} [\mathrm{a} + \mathrm{l}] = \frac{250}{2} [252 + 999] = 156375$