Sum of First n Natural Numbers

We will discuss here how to find the sum of first n natural numbers.

Let S be the required sum.

Therefore, S = 1 + 2 + 3 + 4 + 5 + .................... + n

Clearly, it is an Arithmetic Progression whose first term = 1, last term = n and number of terms = n.

Therefore, S = \(\frac{n}{2}\)(n + 1), [Using the formula S = \(\frac{n}{2}\)(a + l)]

Solved examples to find the sum of first n natural numbers

1. Find the sum of first 25 natural numbers.

Solution:

Let S be the required sum.

Therefore, S = 1 + 2 + 3 + 4 + 5 + .................... + 25

Clearly, it is an Arithmetic Progression whose first term = 1, last term = 25 and number of terms = 25.

Therefore, S = \(\frac{25}{2}\)(25 + 1), [Using the formula S = \(\frac{n}{2}\)(a + l)]

                  = \(\frac{25}{2}\)(26)

                  = 25 × 13

                  = 325

Therefore, the sum of first 25 natural numbers is 325.


2. Find the sum of first 100 natural numbers.

Solution:

Let S be the required sum.

Therefore, S = 1 + 2 + 3 + 4 + 5 + .................... + 100

Clearly, it is an Arithmetic Progression whose first term = 1, last term = 100 and number of terms = 100.

Therefore, S = \(\frac{100}{2}\) (100 + 1), [Using the formula S = \(\frac{n}{2}\)(a + l)]

                  = 50(101)

                  = 5050

Therefore, the sum of first 100 natural numbers is 5050.


3. Find the sum of first 500 natural numbers.

Solution:

Let S be the required sum.

Therefore, S = 1 + 2 + 3 + 4 + 5 + .................... + 500

Clearly, it is an Arithmetic Progression whose first term = 1, last term = 500 and number of terms = 500.

Therefore, S = \(\frac{500}{2}\)(500 + 1), [Using the formula S = \(\frac{n}{2}\)(a + l)]

                  = 225(501)

                  = 112725

Therefore, the sum of first 100 natural numbers is 112725.

Arithmetic Progression




11 and 12 Grade Math

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