Question

For any positive integer n, the sum of the first n positive integers equals n(n+1)2n(n+1)2. what is the sum of all the even integers between 99 and 301 ?

Answer 1

Answer:  The answer is 20200.

Step-by-step explanation:  Given that the sum of first 'n' positive integers is given by

$S_n=\dfrac{n(n+1)}{2}$

and sum of first 'n' even integers is

$S_{en}=n(n+1).$  

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

From 1 to 99, number of even integers is

$\dfrac{99-1}{2}=49,$

and the number of even integers from 1 to 301 is

$\dfrac{301-1}{2}=150.$

Now, sum of first 49 even integers is

$S_{49}=49(49+1)=2450,$

and sum of first 150 even integers is

$S_{150}=150(150+1)=22650.$

Therefore, sum of all even integers between 99 and 301 is given by

$S_{99-301}=S_{150}-S_{49}=22650-2450=20200.$

Thus, the required sum is 20200.

Answer 2

Answer:

The sum of even integers between 99 and 301 is 20200.

Step-by-step explanation:

To find : what is the sum of all the even integers between 99 and 301?

Solution : The even integers between 99 and 301  

100 would be least such integer and 300 would be greatest integer

So series form is 100,102,104,106,.......,300  

Now applying concepts of an arithmetic progression :

Where, First term,  a=100

Common difference,d=102-100=104-102=.....= 2

Last term, l=300

Now the relationship between a,d and I  

$l=a+(n-1)\times d$

where n is number of terms  

$300=100+(n-1)\times 2$

$200=(n-1)\times 2$

$100=n-1$

$101=n$

Now we have to find sum of these 101 terms  

Sum of n terms of an arithmetic progression is

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

$S_{101}=\frac{101}{2}[2(100)+(101-1)2]$

$S_{101}=\frac{101}{2}[200+(100)2]$

$S_{101}=\frac{101}{2}[400]$

$S_{101}=101\times200]$

$S_{101}=20200$

Sum of all even integers between 99 and 301 is 20200.