# How much smaller is the sum of the first 1000 natural numbers than the sum of the first 1001 natural numbers?

Last Updated : 25 Dec, 2023

Numbers are mathematical values or numbers that are used to measure or calculate quantities. It is also represented by digits such as 3, 4, 8, and so on. Other types of numbers include whole numbers, integers, natural numbers, rational and irrational numbers, and so on. A number system is one that incorporates several sorts of numbers such as prime numbers, odd numbers, even numbers, rational numbers, whole numbers, and so on. These numbers can be conveyed in the form of figures as well as words. For example, the numbers 55 and 75, when stated as figures, can also be written as fifty-five and seventy-five.

### Natural Numbers

Natural numbers are a subset of the number system that includes all positive integers from 1 to infinity. Natural numbers are sometimes known as counting numbers since they do not include zero or negative numbers. They are a subset of real numbers that includes only positive integers and exclude zero, fractions, decimals, and negative numbers. Only positive integers, i.e., 1, 2, 3, 4, 5, 6,…, are included in the set of natural numbers.

Sum of Natural Numbers

The sum of n natural numbers is an arithmetic progression in which the sum of n terms is arranged in a sequence with the first term being 1, n being the number of terms, and the nth term being the nth term. Natural numbers are those that begin with one and end with infinity. Except for the number 0, natural numbers contain only whole numbers. The sum of n natural numbers,

[n(n+1)]/2

Example: Find the sum of the first 30 natural numbers.

Solution:

Given, n = 30

The sum of natural numbers formula is:

S = [30(30 + 1)]/2

S = [30(30 + 1)]/2

S = [30(31)]/2

= 930 /2

= 465

### How much smaller is the sum of the first 1000 natural numbers than the sum of the first 1001 natural numbers?

Solution:

To find the difference between both the sum of natural numbers,

For first 1000 natural numbers, [n(n + 1)]/2

Here n = 1000

= [1000(1000 + 1)]/2

= [1000(1001)]/2

= 1001000/2

= 500500

Sum of first 1000 natural numbers (S1) are 500500

Now sum of first 1001 natural numbers are [n(n + 1)]/2 here n = 1001

= [1001(1001 + 1)]/2

= [1001(1002)]/2

= 1003002 / 2

= 501501

Sum of first 1001 natural numbers (S2) are 501501

Now S2 – S1 = 501501 – 500500

= 1001

The sum of the first 1000 natural numbers is smaller by 1001 than the sum of the first 1001 natural numbers.

### Similar Questions

Question 1: Find the sum of the First 100 natural numbers?

Solution:

Here n = 100

Therefore sum of 100 natural numbers are: [n(n + 1)]/2

= [100(100 + 1)]/2

= [100(101)]/2

= 10100/2

= 5050

Question 2: Find the sum of the First 100 even natural numbers?

Solution:

First 100 natural numbers,

As we know there are 100 even natural numbers from 1 to 200.

Therefore n = 100

By using the formula of sum of even numbers Sn = n (n + 1)

Sn = 100 (100 + 1)

= 100 Ã— 101

= 10100

Question 3: Find the sum of the First 200 even natural numbers?

Solution:

First 200 natural numbers,

As we know there are 200 even natural numbers from 1 to 400.

Therefore n=200

By using the formula of sum of even numbers Sn = n (n + 1)

Sn = 200 (200 + 1)

= 200 Ã— 201

= 40200

Question 4: Find the sum of the First 50 Odd natural numbers?

Solution:

First 50 Odd natural numbers,

As we know there are 50 odd natural numbers from 1 to 100.

Therefore n = 50

By using the formula of sum of even numbers Sn = n2

Sn = n2

= 502

= 2500

Question 5: Find the sum of the First 25 Odd natural numbers?

Solution:

First 25 Odd natural numbers,

As we know there are 25 odd natural numbers from 1 to 50.

Therefore n = 25

By using the formula of sum of even numbers Sn = n2

Sn = n2

= 252

= 625

Question 6:  How much smaller is the sum of the first 100 natural numbers than the sum of the first 101 natural numbers?

Solution:

To find the difference between both the sum of natural numbers,

For first 100 natural numbers [n(n + 1)]/2

Here n = 100

= [100(100 + 1)]/2

= [100(101)]/2

= 10100/2

= 5050

Sum of first 1000 natural numbers (S1) are 5050

Now sum of first 101 natural numbers are [n(n + 1)]/2, here n = 101

= [101(101 + 1)]/2

= [101(102)]/2

= 10302 / 2

= 5151

Sum of first 101 natural numbers (S2) are 5151

Now S2 – S1  = 5151 – 5050

= 101

The sum of the first 100 natural numbers is smaller by 101 than the sum of the first 101 natural numbers.