Question

The sum of the digit of all natural number from 1 to 300 is equal to

& A.45150 \\\ & B.3003 \\\ & C.3033 \\\ & D.45679 \\\ \end{aligned}$$

Answer 1

We are given the numbers from 1 to 300 are natural numbers, we are asked to find the sum of these digits. We see that the terms are separated by the same difference. So they form arithmetic sequence so we use sum of n term of arithmetic formula ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ to find the sum of those digit. Here 'a' is first term, n = number of terms, d = difference and ${{S}_{n}}$ denote the sum of n terms, so we apply them to find our required sum.

Complete step-by-step solution:
We are given that, we have numbers from 1 to 300 and we are asked to calculate the sum of these digits.
Firstly we are given natural numbers, we know natural numbers are the numbers that start from 1 and go up forward in possible directions like 1, 2, 3, . . . . . . . . . . .
We are asked to calculate the sum of natural numbers from 1 to 300.
So we have numbers as 1, 2, 3, 4 .......... up to 300.
To find the sum we have more than one way, we will proceed by use of arithmetic progression.
As we can see that, number 1, 2, 3, 4 . . . . . . . . . . . up to 300 are all increasing by same common difference that is 1. So, these numbers are forming an arithmetic progression in which the first term (a) is 1, common difference (d) is 1 and the total number of terms (n) is 300.
Now, as we know, sum of n terms in arithmetic sequence is given as, ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$.
As we have n = 300, a = 1 and d = 1. So we get,
${{S}_{300}}=\dfrac{300}{2}\left( 2\times 1+\left( 300-1 \right)1 \right)$.
Simplifying we get, ${{S}_{300}}=150\left( 2+300-1 \right)$.
Solving further we get ${{S}_{300}}=150\left( 301 \right)$.
Now we get ${{S}_{300}}=45150$.
So the sum of natural numbers from 1 to 300 is 45150.
Hence A is the correct option.

Note: Another method is using sequence of special series in which we know sum of n natural number is given as $\dfrac{n\left( n+1 \right)}{2}$ that is in sequence 1, 2 . . . . . . . . . . . . n.
Sum is given as $1+2+3\ldots \ldots +n=\dfrac{n\left( n+1 \right)}{2}$.
Now we have terms as 1+2+3 . . . . . . . . . +300 so our n is 300. Hence,
$1+2+3\ldots \ldots +300=\dfrac{300\left( 300+1 \right)}{2}$.
Simplifying we get $\Rightarrow 150\times 301$.
Solving we get $\Rightarrow 45150$.
So we get the sum of the numbers from 1 to 300 as 1+2+3 . . . . . . . . . +300 = 45150.