Assertion (A): The sum of first 99 natural numbers is 4950.

Reason (R): The sum of first n natural numbers is $\dfrac{n(n + 1)}{2}$.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Let first n natural numbers : 1, 2, 3, ……., n.

The above sequence is an A.P. with first term = 1, common difference = 1 and number of terms = n.

By formula,

Sum of an A.P. = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$S = \dfrac{n}{2} \times [2 \times 1 + (n - 1) \times 1] \\[1em] = \dfrac{n}{2} \times [2 + n - 1] \\[1em] = \dfrac{n}{2} \times (n + 1).$

So, reason (R) is true.

When n = 99.

The sum of first 99 natural numbers

$= \dfrac{99(99 + 1)}{2}\\[1em] = \dfrac{99 \times 100}{2}\\[1em] = \dfrac{9900}{2}\\[1em] = 4950$

So, assertion (A) is true and, reason (R) is the correct explanation of assertion (A).

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.