Question

find the sum of all natural numbers from 100to200 which are divisible by 4​

Answer 1

Answer:

The sum of natural number between 100 to 200 that are divisible by 4 is $S_{26}=2600$

Step-by-step explanation:

Here required is the sum of all the natural numbers between 100 and 200 which are divisible by 4.

The very next natural number after 100 that is divisible by 4 is 4×25=100

So the series which are exactly divisible by 4 starts from a=100

The nearest natural number divisible by 4 before 200 is 4x50=200

So the series which are exactly divisible by 4 ends with$a_{n}=200$

The series is as below:

100,104,108,..........200

and the difference in each term is d=4

formula for nth term is $a_{n}=a+(n-1)d\\200=100+(n-1)4\\100=(n-1)4\\\frac{100}{4} =n-1\\n=26$

Therefore 200 is 26th term in series and the sum of this series upto 26th term is as below:

$S_{n}=\frac{n}{2} (a+(n-1)d)$

$S_{26} =\frac{26}{2} (100+(26-1)4)\\S_{26}=13 (100+100)\\S_{26}=13 (200)\\S_{26}=2600$

So the sum of natural number between 100 to 200 that are divisible by 4 is $S_{26}=2600$