Question

The sum of all 3 digit numbers:
(a). 494550
(b). 494440
(c). 495550
(d). 495440

Answer 1

Hint: The smallest 3 digit number is 100 and the largest 3 digit number is 999. Find the sum of the series 100, 101, 102,…, 999. This is an arithmetic progression series with common difference 1 and the first term is 100, last term is 999. Use the formula of sum of A.P. series when first and last terms are given. That is:
$\dfrac{n}{2}\left( \text{first term + last term} \right)$

Complete step-by-step answer:

Here, we need to find out the sum of all 3 digit numbers.
We know that the smallest 3 digit number is 100 and the largest 3 digit number is 999. Therefore, we have to add all the numbers starting from 100 up to 999.
So, the problem is as follows:
100 + 101 + 102 +…+ 999
Here we can see that the numbers are in an arithmetic progression series. Where the common difference is 1, the first term is 100 and the last term is 999.
We have a formula for the sum of an arithmetic series when the first and the last terms are given. Than is:
$\dfrac{n}{2}\left( \text{first term + last term} \right)$, n is the number of total terms in the series.
We have $\left( 999-100 \right)+1=900$ terms in our series. Therefore, by applying the above formula the sum of all 3 digit numbers will be:
$=\dfrac{900}{2}\left( 100+999 \right)$
$=450\times 1099=494550$
Hence, the sum of all 3 digit numbers is 494550.
Therefore, option (a) is correct.

Note: Here we can make mistakes while counting the total number of terms. We generally forget to add the last 1. Alternatively we can use the formula of ${{n}^{th}}$ term of an A.P series to count the number of terms. That is : ${{a}_{n}}=a+\left( n-1 \right)d$. Here, the ${{n}^{th}}$ term will be our last term 999, a is the first term and the common difference is 1.