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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 ?

User Galois
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2 Answers

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Answer: The answer is 20200.


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


S_n=(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


(99-1)/(2)=49,

and the number of even integers from 1 to 301 is


(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.

User Bigbad
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Answer:

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

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)* d

where n is number of terms


300=100+(n-1)* 2


200=(n-1)* 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=(n)/(2)[2a+(n-1)d]


S_(101)=(101)/(2)[2(100)+(101-1)2]


S_(101)=(101)/(2)[200+(100)2]


S_(101)=(101)/(2)[400]


S_(101)=101*200]


S_(101)=20200

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

User Bert Bristow
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