Sn = n/2 × (2a + (n-1)d). For natural numbers a=1, d=1. Answer: 5050.

Find the Sum of First 100 Natural Numbers — AP Sum Formula

Question

Find the sum of the first 100 natural numbers.

This is the classic AP problem that Gauss supposedly solved at age 10 — and it shows up in NCERT Class 10, Chapter 5 as a direct application of the sum formula.

Solution — Step by Step

The natural numbers 1, 2, 3, …, 100 form an arithmetic progression. First term $a = 1$ , common difference $d = 1$ , number of terms $n = 100$ .

The sum of $n$ terms of an AP is:

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

We always start with the formula before substituting — this way we don’t lose track of which values go where.

S_{100} = \frac{100}{2} \left[2(1) + (100-1)(1)\right]

= 50 \left[2 + 99\right]

= 50 \times 101

S_{100} = 50 \times 101 = \mathbf{5050}

Answer: The sum of the first 100 natural numbers is 5050.

Why This Works

Every AP has a beautiful symmetry: if you pair the first and last terms, you always get the same sum. Here, $1 + 100 = 101$ , $2 + 99 = 101$ , $3 + 98 = 101$ — and we have 50 such pairs. So the total is $50 \times 101 = 5050$ .

The formula $S_n = \frac{n}{2}[2a + (n-1)d]$ is essentially encoding this pairing logic algebraically. The term $2a + (n-1)d$ is just $a + a_n$ — the sum of the first and last terms.

This is why we can never go wrong if we remember the physical meaning: average of first and last term, multiplied by number of terms.

Alternative Method

There’s a cleaner version of the formula when you already know both the first and last terms:

S_n = \frac{n}{2}(a + l)

Here $a = 1$ , $l = 100$ , $n = 100$ .

S_{100} = \frac{100}{2}(1 + 100) = 50 \times 101 = 5050

Use this form whenever the last term is directly given in the problem — it’s faster and has fewer substitution steps. In problems like “sum of even numbers from 2 to 200”, this form saves 30 seconds.

For natural numbers specifically, there’s a shortcut worth memorising: $S_n = \frac{n(n+1)}{2}$ . This is the same formula simplified for $a = 1, d = 1$ . For $n = 100$ : $\frac{100 \times 101}{2} = 5050$ .

Common Mistake

The most common error here is writing $n = 99$ instead of $n = 100$ — students count “100 natural numbers” as going from 1 to 99 because they subtract 1 somewhere in their head. Count carefully: natural numbers from 1 to 100 are exactly 100 terms. If you’re unsure, use $n = l - a + 1 = 100 - 1 + 1 = 100$ . This cross-check takes 5 seconds and prevents a guaranteed mark loss.

A second slip: using $d = 0$ accidentally after writing the formula, because the numbers “look obvious.” Always write $d = 1$ explicitly before substituting — especially in board exams where step marks matter.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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