The sum of all natural numbers from 100 to 300 which are exactly divisible by 4 and 5 is
Answers
Answer:
Step-by-step explanation:
The numbers which are exactly divisible by 4 and 5 form the following series:
100,120,140,160,.............,300
Hence, a=100,d=20,a^n=300
To find n , we have to apply the formula:
a^n = a+(n-1)d
300=100+(n-1)20
200= (n-1)20
(n-1)=10
n=11
Now, we have to find the sum by applying the formula:
S^n= n/2(a+l)
=11/2(100+300)
=11/2(400)
=2200
Hence, answer is 2200..
Given:
Two natural numbers 100 to 300.
To Find:
The sum of all the natural numbers from 100 to 300 are exactly divisible by 4 and 5 is?
Solution:
The given problem can be solved using the concepts of divisibility rules, and Arithmetic progressions.
1. If a number is divisible by both 4 and 5, then the number is divisible by 20. Hence, if a number is divisible by 20 it will be divisible by both 4 and 5.
2. The list of all the numbers divisible by 20 from 100 to 300 is:
=> 100, 120, 140, 160, 180, 200, 220, 240, 260, 280, 300.
3. The series mentioned above is an Arithmetic Progression with first term (a) = 100, common difference = 20, last term = 300.
=> Let the total number of terms be n,
=> Last term = a + (n-1)d,
=> 300 = 100 + (n-1)20,
=> 200 = 20n - 20,
=> 20n = 220,
=> n = 11.
4. The sum of the first n terms of A.P is given by the formula,
=> Sum = (n/2) (2a + (n-1)xd)
=> Sum = (11/2) (2 x 100 + (11 - 1) x 20),
=> Sum = (11/2) ( 200 + 200),
=> Sum = (11 x 400)/2,
=> Sum = 2200.
Therefore, the sum of all the natural numbers from 100 to 300 that are exactly divisible by 4 and 5 is 2200.