
Find the sum of the first 200 natural numbers.
Answer
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Hint – In order to solve this problem we need to consider numbers from 1 to 200 as an AP and then use the formula of sum of an AP whose last term is given and then you will get the right answer to this problem.
Complete step-by-step answer:
We need to find the sum of the first 200 natural numbers.
So the series of natural numbers is 1, 2, 3,……….200.
Here the first term a is 1.
Common difference d is also 1 clearly.
Last term( l ) is 200.
Number of terms is also 200.
We know the sum of n terms of an AP can be written as ${{\text{S}}_{\text{n}}}{\text{ = }}\dfrac{{\text{n}}}{{\text{2}}}{\text{(a + l)}}$.
On putting the value we get,
$
{{\text{S}}_{200}} = \dfrac{{200}}{2}(1 + 200) \\
{{\text{S}}_{200}} = 100(201) \\
{{\text{S}}_{200}} = 20100 \\
$
Hence, the answer to this question is 20100.
Note – In this problem you need to consider the series from 1 to 200 as an AP and use the formula of sum of 200 terms of an AP. Doing this and solving algebraically you will get the right answer. You can also use the formula of sum as ${{\text{S}}_{\text{n}}}{\text{ = }}\dfrac{{\text{n}}}{{\text{2}}}{\text{(2a + (n - 1)d)}}$ using this also you will get the right answer.
Complete step-by-step answer:
We need to find the sum of the first 200 natural numbers.
So the series of natural numbers is 1, 2, 3,……….200.
Here the first term a is 1.
Common difference d is also 1 clearly.
Last term( l ) is 200.
Number of terms is also 200.
We know the sum of n terms of an AP can be written as ${{\text{S}}_{\text{n}}}{\text{ = }}\dfrac{{\text{n}}}{{\text{2}}}{\text{(a + l)}}$.
On putting the value we get,
$
{{\text{S}}_{200}} = \dfrac{{200}}{2}(1 + 200) \\
{{\text{S}}_{200}} = 100(201) \\
{{\text{S}}_{200}} = 20100 \\
$
Hence, the answer to this question is 20100.
Note – In this problem you need to consider the series from 1 to 200 as an AP and use the formula of sum of 200 terms of an AP. Doing this and solving algebraically you will get the right answer. You can also use the formula of sum as ${{\text{S}}_{\text{n}}}{\text{ = }}\dfrac{{\text{n}}}{{\text{2}}}{\text{(2a + (n - 1)d)}}$ using this also you will get the right answer.
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