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.
1st sequence => (2 + 4 + 6 +…+ 100)
a = 2, d = 4-2 = 2, an = 100
By using the formula,
an = 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
2nd sequence, (5 + 15 + 25 +…+95)
a = 5, d = 15-5 = 10, an = 95
By using the formula,
an = 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