Question

The sum of all natural numbers from 100 to 300 which are exactly divisible by 4 or 5
​

Answer 1

sum of numbers divisible by 4

= 100 + 104 + 108 + ... + 300

total numbers = 51

Sn = n/2[2a + (n – 1)d]

= 51/2[200 + 50 x 4]

= 51/2[400]

= 51 x 200

= 10200

sum of numbers divisible by 5

= 100 + 105 + 110 + 115 + ... + 300

total numbers = 41

Sn = n/2[2a + (n – 1)d]

= 41/2[200 + 40 x 5]

= 41/2[400]

= 41 x 200

= 8200

but above additions contain numbers which are divisible by 20 (LCM of 4 and 5)

to be subtracted.

sum of numbers divisible by 20

= 100 + 120, + 140, + ... + 300

total numbers = 11

= 11/2[200 + 10 x 20]

= 11 x 200

= 2200

required sum = 10200 + 8200 – 2200

= 16200