Question

The sum of natural numbers upto 200 excluding those divisible by 5 is

Answer 1

Answer:

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Answer 2

Answer: 16000

Step-by-step explanation:

Sum of natural numbers upto 200

$1+2+3+4+...... =\dfrac{n(n+1)}{2} =\dfrac{200(200+1)}{2} =\dfrac{200\times 201}{2} =20100$

Now terms divisible by 5

5,10,15,20,................................................,200

It forms an AP where first term a= 5 , common difference d= 5 , last term = 200

$a_n= a+(n-1)d\\\\\Rightarrow 200= 5+(n-1)5\\\\\Rightarrow \dfrac{195}{5} =n-1\\\\\Rightarrow 39= n-1\\\\\Rightarrow n= 40$

Sum of numbers divisible by 5

$S_n= \dfrac{n}{2} (a+l)\\\\\Rightarrow s_{40} = \dfrac{40}{2} (5+200)= 4100$

Sum of natural numbers upto 200 excluding divisible by 5 = Total sum - sum of numbers divisible by 5= 20100 - 4100 = 16000

Hence, the required sum is 16000