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What is the sum of all even natural numbers less than $200$ ?

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1000

10100

9900

10000

answer is C.

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Detailed Solution

The given even natural numbers less than 200 are

2, 4, 6 . . . 19

.
The numbers form an AP where,
First term

a = 2

Common difference

d = 2

a_{n} = 19

We know that

n^{th}

term of an AP is given as,

a_{n} = a + (n - 1) d

Substituting the respective values, we get

\Rightarrow 198 = 2 + (n - 1) \times 2

\Rightarrow 198 - 2 = (n - 1) \times 2

\Rightarrow 196 = (n - 1) \times 2

\Rightarrow \frac{196}{2} = n - 1

\Rightarrow 98 = n - 1

\Rightarrow n = 98 + 1

\Rightarrow n = 99

We know sum of first

n

terms of an AP given as,

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

{\Rightarrow S}_{99} = \frac{99}{2} [2 \times 2 + (99 - 1) 2]

\Rightarrow S_{99} = \frac{99}{2} \times 2 [2 + 98]

\Rightarrow S_{99} = 99 \times 100

\Rightarrow S_{99} = 9900

Therefore, the sum of the required numbers is 9900.
Hence, the correct option is 3.

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