Although a formula is provided in this problem, we can easily solve it using a different formula:

sum = (average)(quantity)

Let’s first determine the average.

In any set of numbers in an arithmetic sequence, we can determine the average using the formula:

(1st number in set + last number in set)/2

Remember, we must average the first even integer in the set and the last even integer in the set. So we have:

(100 + 300)/2 = 400/2 = 200

Next we have to determine the quantity. Once again, we include the first even integer in the set and the last even integer in the set. Thus, we are actually determining the quantity of even consecutive even integers from 100 to 300, inclusive.

Two key points to recognize:

1) Because we are determining the number of “even integers” in the set, we must divide by 2 after subtracting our quantities.

2) Because we are counting the consecutive even integers from 100 to 300, inclusive, we must “add 1” after doing the subtraction.

quantity = (300 – 100)/2 + 1

quantity = 200/2 + 1

quantity = 101

Finally, we can determine the sum.

sum = 200 x 101

sum = 20,200

Note that the reason this is easier than using the formula provided, is that the given formula would have to be applied several times since we don’t want the total of all of the first 300 numbers. We’d have to remember to subtract the sum of the first 99 and divide by two to count only the even numbers. But we’d also have to account for the fact that the first and the last number in the set are both even. So even though a formula is given, it isn’t very easy to use.

Answer is B.
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Hi adkikani

The formula is a simple derivation of AP series that starts from 1 and is consecutive

So the Sum of AP series is \(S_n=\frac{n[2a+(n-1)d]}{2}=\frac{n}{2}\)[First term + Last Term]

if the sequence starts from 1 and is consecutive so your First term=1 and Last Term=number of terms in the series (because sequence is consecutive)

so \(S_n=\frac{n(1+n)}{2}\)

Let's take a simple example. if I have to find sum of numbers from 6 to 10, then I can do that by first finding sum of all Numbers from 1 to 10 and then from this sum subtract sum of numbers from 1 to 5.

In the number line this can be represented as

1...2...3...4...5...6...7...8...9...10. Here First term =1 and last term=10=number of terms in the sequence

As you can clearly see to find sum of numbers from 6 to 10, we simply need to remove sum of numbers from 1 to 5 out of the total sum i.e from 1 to 10.

Now in this question we need to find sum of number from 50 to 150 so it will be Total sum of numbers from 1 to 150 less sum of numbers from 1 to 49

We are given the formula:



We're asked to find sum of all the even numbers between \(99\) and \(301\)....



We can factor out \(2\) from this sum:



In the brackets we have sum of numbers from \(50\) to \(150\). We can use the given formula for the sum of \(1\) to \(150\) and subtract the sum of \(1\) to \(49\) to find the total of this sum.






The final answer is .

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

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

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

To find the solution, treat this like any other consecutive integer problem. To find the sum of any consecutive integers, multiply the average of those integers times the number of integers.

So first, to find the average of the integers, we know that the lower limit is 100 and the upper limit is 300, since we are only looking for even integers. To find the average, (300 + 100) / 2, which gives us 200.

Next, find the total number of integers. We know that 100 is the 50th even integer and that 300 is the 150th even integer. We subtract 150 - 50 to get 100 even integers, but must add 1 to the total since we are looking for an INCLUSIVE number.

So, we know the average of the integers in this range is 200 and the number of integers in the range is 101. Follow the principles of consecutive integers to find the answer: SUM = (avg) ( # of integers).

This gets us 20,200 = 200 * 101.