Can the same approach (refer to NCERT page no. 183) be used to find the sum of $1 + 2 + 3 + \dots + 100$?                                                                                                                                                                                                                                   [Page No. 184]

The sum of the first 100 natural numbers can be found using the forward-and-backward pairing method.

$$\text{Let, } S = 1 + 2 + 3 + ... + 100.$$

Writing the sum in reverse order,

$$S = 100 + 99 + 98 + ... + 1.$$

Adding the two equations term by term,

$$2S = (1 + 100) + (2 + 99) + (3 + 98) + ... + (100 + 1).$$

Each pair sums to 101, and there are 100 such pairs. Therefore,

$$2S = 101 \times 100 = 10100.$$

$$\text{Hence, } S = \frac{10100}{2} = 5050.$$

$$\text{Thus, } 1 + 2 + 3 + ... + 100 = 5050$$

This method works for any arithmetic sequence and leads to the formula

$$1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$$