What is the sum of first $30$ natural numbers?

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465

410

420

455

answer is A.

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Detailed Solution

The given natural numbers are 1,2,3……
The natural number form an AP with first term 1 and common difference 1.
So, we have,

a = 1, d = 1

.
We know that sum of first n terms of an AP is given as,

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

.
Putting the value of a = 1 and d = 1, we get

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

\Rightarrow S_{n} = \frac{n}{2} [2 + n - 1]

\Rightarrow S_{n} = \frac{n (n + 1)}{2}

[1]
Substituting

n = 30

in [1], we get

S_{100} = \frac{30 (30 + 1)}{2}

\Rightarrow S_{100} = 465

Therefore, the sum of the first 30 natural numbers is 465.
Hence, the correct option is 1.

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