Question

For any positive integer $n$, the sum of the first $n$ positive integers equals $\frac{n(n+1)}{2}$. What is the sum of all the even integers between $99$ and $301$?

• A. $10100$
• B. $20200$
• C. $22650$
• D. $40200$
• E. $45150$

Answer 1

The correct option is B $20200$

We need to find sum of all even integers between $99$ to $301$.

First even integer will be $100$ and last will be $300$.

Common difference between terms will be $2$.

Therefore, $300=100+(n-1)2$

$\Rightarrow 200=2n-2$

$\Rightarrow n=101$

Thus $\text{sum}=\frac{101}{2}(100+300)$

$=\frac{101}{2}\times 400$

$=101\times 200$

$=20200$