Answer:

The natural numbers which are divisible by 2 or 5 are:

2 + 4 + 5 + 6 + 8 + 10 + … + 100 = (2 + 4 + 6 +…+ 100) + (5 + 15 + 25 +…+95)

(2 + 4 + 6 +…+ 100) + (5 + 15 + 25 +…+95) are AP with common difference of 2 and 10.

1^st sequence => (2 + 4 + 6 +…+ 100)

a = 2, d = 4-2 = 2, a_n = 100

By using the formula,

a_n = a + (n-1)d

100 = 2 + (n-1)2

100 = 2 + 2n – 2

2n = 100

n = 100/2

n = 50

S = n/2 (2a + (n-1)d)

S = 50/2 (2(2) + (50-1)2)

S = 25 (4 + 49(2))

S = 25 (4 + 98)

S = 2550

2^nd sequence, (5 + 15 + 25 +…+95)

a = 5, d = 15-5 = 10, a_n = 95

By using the formula,

a_n = a + (n-1)d

95 = 5 + (n-1)10

95 = 5 + 10n – 10

10n = 95 +10 – 5

10n = 100

n = 100/10

n = 10

S = n/2 (2a + (n-1)d)

S = 10/2 (2(5) + (10-1)10)

S = 5 (10 + 9(10))

S = 5 (10 + 90)

S = 500

∴ The sum of the numbers divisible by 2 or 5 is: 2550 + 500 = 3050