Question -
Answer -
The natural numberswhich are divisible by 2 or 5 are:
2 + 4 + 5 + 6 + 8 + 10+ … + 100 = (2 + 4 + 6 +…+ 100) + (5 + 15 + 25 +…+95)
Now, (2 + 4 + 6 +…+100) + (5 + 15 + 25 +…+95) are AP with common difference of 2 and 10.
So, for the 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
= 50
So now, S = n/2 (2a +(n-1)d)
= 50/2 (2(2) +(50-1)2)
= 25 (4 + 49(2))
= 25 (4 + 98)
= 2550
Again, for the 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
= 10
So now, S = n/2 (2a +(n-1)d)
= 10/2 (2(5) +(10-1)10)
= 5 (10 + 9(10))
= 5 (10 + 90)
= 500
∴ The sum of thenumbers divisible by 2 or 5 is: 2550 + 500 = 3050