Question

sum of all natural number between 100 and 400 which are divisible by 3 is?​

Answer 1

Answer:

Step-by-step explanation:

1st number divisible by 3 between 100 to 400 is 102

and last number is  399

102, 105 ,108,................................., 399

it became an A.P.

in which a = 102 and d = 3

$a_{n} = a + (n-1 ) .d$

399 = 102 + (n - 1) 3

from here n = 100

$S_{n} = \frac{n}{2}[a + a_{n}]$

   = 100/2 [102 + 399]

   = 50 [508]

  =  25400