Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.

$\dfrac{250}{9} = 27\dfrac{7}{9} \text{ and } \dfrac{1000}{9} = 111\dfrac{1}{9}$.

The numbers which are divisible by 9 between 250 and 1000 are,

= 28 × 9, 29 × 9, 30 × 9, …………, 111 × 9.

= 252, 261, 270, ………., 999.

The above sequence is an A.P. with common difference = 9 and first term = 252 and last term = 999.

Let n be no. of terms,

∴ a_n = a + (n - 1)d

⇒ 999 = 252 + (n - 1)9

⇒ 999 = 252 + 9n - 9

⇒ 999 = 9n + 243

⇒ 999 - 243 = 9n

⇒ 9n = 756

⇒ n = 84.

$S = \dfrac{n}{2}(a + l) \\[1em] = \dfrac{84}{2} \times (252 + 999) \\[1em] = 42 \times 1251 \\[1em] = 52542.$

Hence, sum = 52542.