Question

find the sum of all natural numbers between 400 and 600 which are divisible by 9

Answer 1

The numbers lying between 400 and 600 which are divisible by 9 are.

= > 405, 414,423, 432,441.... 594.

Here, First term, a = 405.

Last term, l = 594.

Common difference d = 414 - 405 = 9.

Let the number of terms of the A.P be n.

= > an = a + (n - 1) * d

= > 594 = 405 + (n - 1) * 9

= > 594 = 405 + 9n - 9

= > 594 = 396 + 9n

= > 198 = 9n

= > n = 22.


Now,

Sum of terms sn = n/2(a + l)

                            = 22/2(405 + 594)

                            = 11(999)

                            = 10989.



Therefore the required sum = 10989.


Hope this helps!