Question

find the sum of all natural numbers between 100 and 200 which are completely divisible by 5​

Answer 1

There are 19 numbers between 100 and 200 which are divisible by 5. Those numbers are 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175, 180, 185, 190, 195...

Ap = 105,110, 115, 120, 125..195

first term , a= 105

common difference , d=5

last term , l = 195

last term = a +(n-1)d

195 = 105 +(n-1)5

195-105 = (n-1)5

90 = (n-1)5

n-1 = 90/5

n-1= 18

n= 18+1

n=19

Sn =

$= \frac{n}{2} (a \: + l) \\ = \frac{19}{2} (105 +195) \\ = \frac{19}{2} (300) \\ = 19 \times 150 \\ = 2,850$

the sum of all natural numbers between 100 and 200 which are completely divisible by 5

= 2,850

Ans