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
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