Question

The sum of all natural numbers from 100 to 300 which are exactly divisible by 4 and 5 is

Answer 1

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..

Answer 2

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.