Question

The sum of the first 100 natural numbers is greater than the sum of the even numbers between 1 and 100 (including 100) by _____.

• A. 2200
• B. 2400
• C. 2500
• D. 2700

Answer 1

The correct option is C

2500

The first 100 natural numbers form an AP with the first term, a = 1 and common difference, d = 1.

Therefore, the sum of the first 100 terms of this AP is given by using the values of a, d and n as 1, 1 and 100 respectively, in the formula:

$S_n=\left(\frac{n}{2}\right)[2a+(n-1\left)\right]d]$

$\therefore$ Sum of the first 100 natural numbers
$=\frac{100}{2}[2+99]$

$=\frac{100}{2}\left[101\right]$

$=50\times 101$

$=5050$

The even numbers between 1 and 100 are 2, 4, 6, 8, ...., 100 which is also an AP with a = 2, common difference, d = 2.
We also know that there are 50 even numbers between 1 and 100, which means n = 50.

$\therefore$The sum of even numbers from 1 to 100
$=\frac{50}{2}(2\times 2+49\times 2)$

$=\frac{50}{2}(4+98)$

$=\frac{50}{2}\left(102\right)$

$=50\times 51$

$=2550$

Hence, the difference is

$5050-2550=2500$