Hi neerajeai, please note that the n(n+1)/2 shortcut formula is only applicable if the starting point is 1. Anytime you want to find the number of terms between two given numbers you should use the general formula ((first - last) / frequency) + 1. You can also multiply by the average at the end to get the sum.

In your case it is (((301-99)/2) + 1) * 200 = 102 * 200 = 20400. Answer choice C with a presumed typo in the unit digit?

Hope this helps!
-Ron
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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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niks18

I was confused if the second formula was applied only when my sequence is starting from 1.
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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

Here's one approach.

We want 100+102+104+....298+300
This equals 2(50+51+52+...+149+150)
From here, a quick way is to evaluate this is to first recognize that there are 101 integers from 50 to 150 inclusive (150 - 50 + 1 = 101)

To evaluate 2(50+51+52+...+149+150), let's add values in pairs:

....50 + 51 + 52 +...+ 149 + 150
+150+ 149+ 148+...+ 51 + 50
...200+ 200+ 200+...+ 200 + 200

How many 200's do we have in the new sum? There are 101 altogether.
101 x 200 = 20,200

Answer: B


Approach #2:

From my last post, we can see that we have 101 even integers from 100 to 300 inclusive.

Since the values in the set are equally spaced, the average (mean) of the 101 numbers = (first number + last number)/2 = (100 + 300)/2 = 400/2 = 200

So, we have 101 integers, whose average value is 200.
So, the sum of all 101 integers = (101)(200)
= 20,200
= B


Approach #3:

Take 100+102+104+ ...+298+300 and factor out the 2 to get 2(50+51+52+...+149+150)
From here, we'll evaluate the sum 50+51+52+...+149+150, and then double it.

Important: notice that 50+51+.....149+150 = (sum of 1 to 150) - (sum of 1 to 49)

Now we use the given formula:
sum of 1 to 150 = 150(151)/2 = 11,325
sum of 1 to 49 = 49(50)/2 = 1,225

So, sum of 50 to 150 = 11,325 - 1,225 = 10,100

So, 2(50+51+52+...+149+150) = 2(10,100) = 20,200

Answer: B

Cheers,
Brent
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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 .

So I know, that we must use the first and last digits of even numbers, if we are calculating for a sum of evens..

But I tried it this way, and still got the answer..
Would like to know, if it's an acceptable method, or its wrong, and I shouldn't use it ?

if we use first and last terms as given, i.e. First term= 99 and last term = 301
Subtract these two numbers, to get the total number of terms..
301-99 = 202
Note this includes odd and evens both

for even, we must divide it by 2.
therefore, we get a 202/2 = 101 terms...

AS you can see, nothing is added or subtracted for inclusive, exclusive..

now, 101 * avg of terms, will give us the sum.

avg = 301+99/2 = 400/2 = 200

101*200 = 20,200
option B

Please correct me, if you think this is a wrong approach and we should not use it

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

In my case I remember the formula of arithmetic progression: Sum of n terms = n/2*(a1+an)
In this case:

Interval: 100-102-104-106........-296-298-300

The difference (distance) between each number is 2:
- If I do (102-100)/2=1
- If I do (104-100)/2=2
- If I do (106-100)/2=3
As we can see, if we do this, every term is in the (n-1) position... so (300-100)/2=100 will be in the n-1 position, then... the number 300 is in the position (100+1=101) of the serie.

Now we apply the formula: Sum of n = n/2*(a1+an)
Sum of n = 101/2*(100+300)=20.200

I Hope this solution could help someone

Thanks!

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.

A great trap choice here would have been 20400 IMO.

In fact, I got this right using the formula : \(\frac{First + Last}{2} * \frac{Last - First}{2} + 1\)

I substituted and got \(200 * \frac{301-99}{2} + 1 = 20400\)

I now know why I am wrong! Thank GMATCLUB

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
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This can also be solved with sequence formula :
Sn = n/2 [ 2a+ (n-1)d]
We have n = number of even integers between 99 & 301,
thus, n = 101,
a = first term = 100,
d = difference between the terms = 2 ( as we want even terms),

Sn = 101/2 * [ 2*100 + (100)*2]
= 101/2 * (400)
= 101*200 = 20200
Thus the correct answer choice is 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

Solution:
Even integer between 99 and 301 are: 100, 102, 104, . . ., 300.
In any set of consecutive integers the sum = Average × Number of terms
Average of the set = (Largest + Smallest)/2 = (300 + 100)/2 = 200
Number of terms = (Largest – Smallest)/2 + 1 = (300 – 100)/2 + 1 = 101
Therefore, the sum of all the even integers = 200 × 101 = 20,200.
Answer: (B) 20,200

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Need to find sum of even terms, 100 +102+104+....300

Find number of even terms: 300 = 100 + (n-1)2
n = 101
Sum of even terms = n/2 [2a +(n-1)2] = 101/2 [200 + (100)2] = 101 * 200 = 20,200

sum of even from 99 to 301
300*302/4 - 98*100/4
22650-2450 ; 20200
option B

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Easiest way

100 is the first even number
101 to 200 , there are 50 even numbers
201-300 , there are another 50 even numbers

so the total even numbers are 50+50+1=101
Also note that all the even consecutive numbers are in AP.
Therefore the sum is given as : SUM= (N*Mean of the AP)
N=101 Mean of AP=first term+last term/2 ; first term in this case is 100 and last term is 300 , therefore the mean is 200

Consequently the sum is 101*200=20,200

hope this helps .

My take: the answers choices are far apart enough to estimate. No need to calculate to the exact.

Sum (first 301 numbers) = (301+302)/2 => est. 300*150=45000
Sum (first 98 numbers) = (98+99)/2 => est. 100*50=5000

Therefore sum (99 to 301) is estimated to be around 40,000. But since we are asked about the even numbers in between, simply divide it by 2 to get around 20,000 => B. 20,200