How to find the sum of all natural numbers from 100 to 300 which are not exactly divisible by 4 and 5
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Answer:
Step-by-step explanation:
sum of numbers divisible by 4
= 100 + 104 + 108 + ... + 300
total numbers = 51
Sn = n/2[2a + (n – 1)d]
= 51/2[200 + 50 × 4]
= 51/2[400]
= 51 × 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 × 5]
= 41/2[400]
= 41 × 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 × 20]
= 11 × 200
= 2200
required sum = 10200 + 8200 – 2200
= 16200
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