Question

What is the sum of all natural numbers from 1 to 100?
(a) 5050
(b) 50
(c) 4550
(d) 5150

Answer 1

Hint: We solve this problem by using the arithmetic progression.
The general representation of an A.P is given as
\[a,a+d,a+2d,......\]
Where, \[a\] is the first term and \[d\] is the common difference.
The sum of \[n\] terms of the A.P is given as
\[{{S}_{n}}=\dfrac{n}{2}\left( a+{{a}_{n}} \right)\]
Where \[{{a}_{n}}\] is the last term of the series.

Complete step by step answer:
We are asked to find the sum of all natural numbers from 1 to 100
We know that the series of natural numbers is
1, 2, 3, 4, ……, 100
Here, we can see that it forms an A.P
We know that the natural numbers are in A.P with first term 1 and last term as 100
We know that the sum of \[n\] terms of the A.P is given as
\[{{S}_{n}}=\dfrac{n}{2}\left( a+{{a}_{n}} \right)\]
Where \[{{a}_{n}}\] is the last term of the series.

By using the above formula to natural numbers from 1 to 100 then we get
\[\begin{align}
  & \Rightarrow {{S}_{100}}=\dfrac{100}{2}\left( 1+100 \right) \\
 & \Rightarrow {{S}_{100}}=50\times 101 \\
 & \Rightarrow {{S}_{100}}=5050 \\
\end{align}\]
Therefore we can conclude that the sum of natural numbers from 1 to 100 is 5050
So, option (a) is correct answer.

Note:
We can solve this problem in another method.
We have the direct formula for the sum of first \[n\] natural numbers as
\[{{S}_{n}}=\dfrac{n\left( n+1 \right)}{2}\]
We are asked to find the sum of natural numbers 1 to 100 which is the sum of the first 100 natural numbers.
By using the above formula we get the required sum as
\[\begin{align}
  & \Rightarrow {{S}_{100}}=\dfrac{100\left( 100+1 \right)}{2} \\
 & \Rightarrow {{S}_{100}}=50\times 101 \\
 & \Rightarrow {{S}_{100}}=5050 \\
\end{align}\]
Therefore we can conclude that the sum of natural numbers from 1 to 100 is 5050
So, option (a) is correct answer.