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

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1000

10100

9900

10000

answer is D.

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Detailed Solution

The given odd natural numbers less than 200 are

1, 3, 5 . . . 199

.
The numbers form an AP where,
First term

a = 1

.
Common difference

d = 2

a_{n} = 199

.
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 199 = 1 + (n - 1) \times 2

\Rightarrow 199 - 1 = (n - 1) \times 2

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

\Rightarrow \frac{198}{2} = n - 1

\Rightarrow n = 99 + 1

\Rightarrow n = 100

We know sum of first

n

terms of an AP given as

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

\Rightarrow S_{100} = \frac{100}{2} [2 \times 1 + (100 - 1) 2]

\Rightarrow S_{100} = \frac{100}{2} \times 2 [1 + (100 - 1)]

\Rightarrow S_{100} = 100 \times [1 + 100 - 1]

\Rightarrow S_{100} = 100 \times 100

\Rightarrow S_{100} = 10000

Therefore, the sum of the required numbers is 10000.
Hence, the correct option is 4.

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