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The sum of natural numbers upto 200 excluding those divisible by 5 is:
- a)20100
- b)4100
- c)16000
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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The sum of natural numbers upto 200 excluding those divisible by 5 is:...
Given: sum of natural numbers upto 200 excluding those divisible by 5.
To find: the sum of the numbers.
Solution:
We can find the sum of natural numbers upto 200 and then subtract the sum of numbers divisible by 5 from it.
Sum of natural numbers upto 200:
Using the formula,
Sum = (n/2) × (first number + last number)
where n is the number of terms.
Here, n = 200 and first number = 1, last number = 200
Sum of natural numbers upto 200 = (200/2) × (1 + 200) = 20100
Sum of numbers divisible by 5:
Using the formula,
Sum = (n/2) × (first number + last number)
where n is the number of terms.
Here, first number = 5, last number = 200, common difference = 5
The formula for the nth term of an arithmetic progression is:
an = a1 + (n-1)d
Using this formula, we can find the number of terms in the sequence of numbers divisible by 5.
200 = 5 + (n-1)5
195 = 5n
n = 39
Number of terms = 39
Sum of numbers divisible by 5 = (39/2) × (5 + 200) = 3970
Sum of natural numbers upto 200 excluding those divisible by 5:
Sum = Sum of natural numbers upto 200 - Sum of numbers divisible by 5
Sum = 20100 - 3970
Sum = 16130
Therefore, the sum of natural numbers upto 200 excluding those divisible by 5 is 16130. Hence, option (c) is correct.
To find: the sum of the numbers.
Solution:
We can find the sum of natural numbers upto 200 and then subtract the sum of numbers divisible by 5 from it.
Sum of natural numbers upto 200:
Using the formula,
Sum = (n/2) × (first number + last number)
where n is the number of terms.
Here, n = 200 and first number = 1, last number = 200
Sum of natural numbers upto 200 = (200/2) × (1 + 200) = 20100
Sum of numbers divisible by 5:
Using the formula,
Sum = (n/2) × (first number + last number)
where n is the number of terms.
Here, first number = 5, last number = 200, common difference = 5
The formula for the nth term of an arithmetic progression is:
an = a1 + (n-1)d
Using this formula, we can find the number of terms in the sequence of numbers divisible by 5.
200 = 5 + (n-1)5
195 = 5n
n = 39
Number of terms = 39
Sum of numbers divisible by 5 = (39/2) × (5 + 200) = 3970
Sum of natural numbers upto 200 excluding those divisible by 5:
Sum = Sum of natural numbers upto 200 - Sum of numbers divisible by 5
Sum = 20100 - 3970
Sum = 16130
Therefore, the sum of natural numbers upto 200 excluding those divisible by 5 is 16130. Hence, option (c) is correct.
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The sum of natural numbers upto 200 excluding those divisible by 5 is:...
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The sum of natural numbers upto 200 excluding those divisible by 5 is:a)20100b)4100c)16000d)None of theseCorrect answer is option 'C'. Can you explain this answer?
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