For any positive integer n, the sum of the first n positive integers

Since it is the even numbers, you can change the set to 100 to 300.

The average of the two is 200.

The amount of even integers is: ((300-100)/2)+ 1 = 101

200 * 101 = 20,200

The sum of the first n terms of an arithmetic pogression is : n/2 *(2a + (n-1)d)

where a is the first term and d is the difference between consecutive terms.


we are to find out the sum of all even integers between 99 and 301

The even numbers look like 100,102,104....300

a=100
n=101
d=2

placing these values in the formula n/2 *(2a + (n-1)d)

sum of even integers = 101/2 * (2*100 + (101-1)*2)
=20200

Choice B.




Alternative method:


The sum of first n positive integers is n(n+1)/2

The sum of first 301 integers is 301*302/2=301*151=45451
The sum of first 99 integers is 99*100/2=99*50=4950

The sum of integers between 99 and 301 = 45451-4950=40501

The above contains all odd and even integers between 99 and 301 --- odd + even = 40501 ------->eqn 1



consider the following series:

numbers in series number of odd numbers sum of odd numbers(so) sum of even numbers(se) so-se
123 2 4 2 2
12345 3 9 6 3
1234567 4 16 12 4


You can see that in the first n integers, the difference of the sum of the odd numbers - sum of even numbers = number of odd numbers in the set.

between 1 and 301 we have 151 odd numbers
between 1 and 99 we have 50 odd numbers

difference of sum of odd numbers and sum of even numbers between 99 and 301 = 151-50=101 ----------> eqn 2

consider eqn 1 and eqn 2

odd+even=40501
odd-even=101

subtract 2 from 1

2even=40400
sum of even numbers between 99 and 301=20200

Choice B

Approach 3

you know that sum of odd + even between 99 and 301 is 40501 (same from previous approach)

The answer should be roughly half of this number --- 20250
closest is B.


Would love to see a simpler more elegant solution. It will save time for us.

the sum of 1st even numbers = n(n+1)
the sum of first even numbers till 301 is 150*(150+1) =22650
the sum of first even numbers till 99 is 49*(49+1) =2450

22650-2450=20200

this method is not perfect. I wrote it just to show one more method
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Easier way would be using sum of all n even integers = n * (n+1)

Hence from 1-301 there are 150 even integers hence sum = 150 * 151 = 22650
From 1- 99 there are 49 even integers ( - 1 as 100 is not part of this) = 49 * 50 = 2450

So sum from 99-301 = 22650 -2450 = 20200

This is another way to tackle this question:

It is given that the sum of the first n positive integers is n(n+1)/2. I didn't really use this formula.

I remembered this formula instead (and I think you should remember it too!):

The sum of the elements in any evenly spaced set is: (mean)*(# of terms)

Steps:
1. First, find the mean
The questions asked to find the sum of all the EVEN integers between 99 and 301. So that means we need to look at EVEN numbers only. So really the range we are looking at is between 100 and 300. So the mean of these evenly spaced numbers between 100 and 300 is 200. --> this is the mean (no calculation necessary here because it's an evenly spaced range so its very straight forward to find the mean; just get the middle number!)

2. Second, find the # of EVEN terms.
We stated in step 1 that the EVEN number range is between 100 and 300. So here we can utilise this formula:
[(last term -first term)/2] +1

which comes to:
[(300-100)]/2 + 1
= (200/2) +1
= 100+1
= 101 --> this is the # of EVEN terms in the range 100 and 300

Coming back to this: The sum of the elements in any evenly spaced set is: (mean)*(# of terms)

(mean)*(# of terms)
=200* 101
=20200

Answer B

Sum of all the even integers between 99 and 301:

The first even integer will be 100 and the last will be 300.

The common difference will be 2.

Number of terms: 300 = 100 + (n-1)2

=> 200 = 2n-2

=> 202 = 2n

=> n = 101

Sum: \(\frac{n}{2}\)[first term + last term]

=> \(\frac{101}{2}\)[100+300]

=> \(\frac{101}{2}\) * 400

=> 101 *200

=> 20200



Answer B

Video solution from Quant Reasoning:
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To find the sum of all the even integers between 99 and 301, we first need to determine the number of even integers in that range.

The largest even integer within the given range is 300, and the smallest even integer is 100. The set of even integers between 99 and 301 can be expressed as {100, 102, 104, ..., 298, 300}.To find the number of terms in this arithmetic sequence, we can use the formula for the nth term of an arithmetic sequence:

an = a1 + (n - 1)d

In this case, a1 = 100 (the first term), d = 2 (the common difference), and we need to find n.

300 = 100 + (n - 1)2
200 = 2n - 2
202 = 2n
n = 101

Therefore, there are 101 even integers between 99 and 301.

Now, we can calculate the sum using the formula for the sum of an arithmetic series:

Sn = (n/2)(a1 + an)
Sn = (101/2)(100 + 300)
Sn = (101/2)(400)
Sn = 101 * 200
Sn = 20,200

The sum of all the even integers between 99 and 301 is 20,200.

Therefore, the answer is option B.

In approach no. 1 - what would have been the difference if they wouldnt have mentiond 'even' numbers?

If we were asked to find the sum of the integers between 99 and 301, inclusive, we can utilize the fact that the sum of the elements in any evenly spaced set is given by: \(Sum=\frac{a_1+a_n}{2}*n\), the mean multiplied by the number of terms. Therefore, the answer would be (99 + 301)/2*203 = 40,600.
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Given: For any positive integer n, the sum of the first n positive integers equals n(n+1)/2.
Asked: What is the sum of all the even integers between 99 and 301?

Sum of all even integers between 99 and 301 = 100 + 102 + .... + 300 = (2 + 4 + .... + 300) - (2 + 4 + ... + 98) =
= 2 (1+ 2+ .... + 150) - 2 (1+2+... + 49) = 150*151 - 49*50 = 50 (453 - 49) = 50*404 = 20200

IMO B

My method was similar to approach #1 here, please let me know if this is wrong

No. of terms = 300-100 + 1 = 201 terms

Sum of 201 terms in this sequence = (201(202)) / 2 = 20200

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