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sum of all even numbers from 99 to 301

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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sum of all even numbers from 99 to 301

by MG368 » Sat Nov 21, 2009 7:50 am
I always get stuck on these type of question. When i look at the explanation it seems simple enough but just can't figure it out.Does anyone have a quick way to solve this type of question which may have variations such as:

-Sum of all even numbers from 1 to 199
-sum of all odd number from 100 to 300
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by imperialhuntress » Mon Nov 23, 2009 2:45 pm
I don't know what explanation you saw, but I solve these problems this way.

When you are dealing with the sum of even or odd integers, consider what nth even/odd integer it is. Then apply a simple formula. For example, when you deal with sum of all even integers from 1 to 99, you can basically rephrase it to say you want the sum of the 1st to 49th EVEN integer. The formula after that would just be n(n+1). So the answer to your first question would be 49(50).

Your second question asks the sum of all odd numbers from 100 to 300. Rephrased in another way, you are asking the sum of the 150th ODD integer minus the sum of the 50th ODD integer. The formula for the sum of the Nth ODD integer would just be n-squared. So the answer to your question would be 150-squared minus 50-squared.

Hope this makes sense. I had a hard time figuring this out but through trial and error playing with numbers figured this out...
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by adam15 » Mon Nov 23, 2009 3:07 pm
the answer for the number of even integer from 1 to 199, (198+2-2)/2=99
for the second one from 100 to 300 how many odd number:
(300+2-100)/2=(202)/2=101 the number of even numbers
then the number of odd number is 200-101=99 odd number.
general rule find the number of even numbers then substract the number from the number of element you have;
number=(last+2-first)/2
number of odd= total number of element-number of even numbers

I hope this will help
sorry for prvious post
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by palvarez » Wed Nov 25, 2009 1:22 pm
MG368 wrote:I always get stuck on these type of question. When i look at the explanation it seems simple enough but just can't figure it out.Does anyone have a quick way to solve this type of question which may have variations such as:

-Sum of all even numbers from 1 to 199
-sum of all odd number from 100 to 300
The trap is this: 99 and 301 are odds.

Reformulate the question: sum of all evens from 100 to 300, inclusive.

Average = (100+300)/2 = 150
number of evens = [(300-100)/2]+1 = 101

101*150 is the answer.
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by mp2437 » Wed Nov 25, 2009 2:49 pm
palvarez wrote:
MG368 wrote:I always get stuck on these type of question. When i look at the explanation it seems simple enough but just can't figure it out.Does anyone have a quick way to solve this type of question which may have variations such as:

-Sum of all even numbers from 1 to 199
-sum of all odd number from 100 to 300
The trap is this: 99 and 301 are odds.

Reformulate the question: sum of all evens from 100 to 300, inclusive.

Average = (100+300)/2 = 150
number of evens = [(300-100)/2]+1 = 101

101*150 is the answer.
He meant well. Average of 100 and 300 is 200, not 150. 200 * 101 is the answer.
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by nikilnikil » Wed Nov 25, 2009 6:40 pm
you can easily solve such problems using "AP-arithmetic progression" formula, if you are not able to solve using the above mentioned method.

Tn=a+(n-1)d

Tn=last term
a= first term
n=number of terms
d= difference between terms

In the above case series is something like
2,4,6,..........198

So a=2
Tn=198
d=2

we have to first find n using the above formula

198=2+(n-1)2
196=(n-1)2
196/2=n-1
n= 196/2+1 = 98+1 = 99


now calculate the sum using the below formula

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

where
Sn= sum of n terms

plug in the above numbers in the formula. you are done.

~Nik
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by sk02 » Wed Nov 25, 2009 9:46 pm
Another way to approach these questions is to start on either end of the range and work your way to the middle.

1) Sum of all even numbers from 1 to 199
Consider the first even number greater then 1 and first even number smaller then 199, so 2 + 198 =200
Then the second even number greater then 1 and second even number smaller then 199, so 4 + 196 =200
Doing the same for each even number or rather multiple of 2, we can continue till 98 + 102 =200.
As we know there will be 98/2 such pairs, the sum must equal (98/2)*200.
BUT we also need to remember that there is one even number with no pair, that is 100, so we add this to the above to arrive at sum = ((98/2)*200)+100 = 9900

2) Sum of all odd number from 100 to 300
Again smallest odd and largest odd in this range 101+299=400, 103+297=400 ... 199 + 201=400
number of such pairs: 50, and in this case unpaired is 0, so we can get the sum by 400*50=20000

In summary, first figure out the sum of the pair, and the number of times these pairs repeat. And finally account for any unpaired integer within the specified range.
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by sreak1089 » Fri Nov 27, 2009 6:22 am
Sum of all even numbers from 1 to 199

== 2 + 4 +6+ 8 + ... + 198 == 2 (1 + 2 + 3 + ... 99) = 2 * (99 * 100) / 2 = 9900

Sum of all odd numbers from 100 to 300
== 101 + 103 + ... + 299

Equally Spaced sets ==> Mean = (101 + 299)/2 = 200
No. of terms = (299 - 101)/2 + 1 = 100

Sum of the series = 200 * 100 = 20000
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by Birgit Anne » Mon Nov 30, 2009 8:56 am
sreak1089 wrote:Sum of all even numbers from 1 to 199

== 2 + 4 +6+ 8 + ... + 198 == 2 (1 + 2 + 3 + ... 99) = 2 * (99 * 100) / 2 = 9900

Sum of all odd numbers from 100 to 300
== 101 + 103 + ... + 299

Equally Spaced sets ==> Mean = (101 + 299)/2 = 200
No. of terms = (299 - 101)/2 + 1 = 100

Sum of the series = 200 * 100 = 20000
Can you explain me why you have to add 1 in order to get the no. of terms?
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by ssuarezo » Wed Dec 02, 2009 1:16 pm
Birgit Anne wrote:
sreak1089 wrote:Sum of all even numbers from 1 to 199

== 2 + 4 +6+ 8 + ... + 198 == 2 (1 + 2 + 3 + ... 99) = 2 * (99 * 100) / 2 = 9900

Sum of all odd numbers from 100 to 300
== 101 + 103 + ... + 299

Equally Spaced sets ==> Mean = (101 + 299)/2 = 200
No. of terms = (299 - 101)/2 + 1 = 100

Sum of the series = 200 * 100 = 20000
Can you explain me why you have to add 1 in order to get the no. of terms?
Hi:

It's part of the formula. I think sreak1089's is the easiest way.

This applies for any even or odd number:

No. of terms = (Last-first)/2 +1 .. explanation:

Last and first depend on if we are asked odd or even numbers in a given set ..
Divided by 2 because we take a number every 2 numbers, even if odd or even numbers.
If we were asked multiples of 5, it would be divided by 5, with first and last multiples of 5 within the given range.

Silvia
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by kabirmohammed » Wed Dec 02, 2009 3:23 pm
the total number of numbers from 100 to 300 is 201.....
thus, total odd nos = (201-101) =100
correct me if Iam wrong
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by ssuarezo » Thu Dec 03, 2009 8:05 am
kabirmohammed wrote:the total number of numbers from 100 to 300 is 201.....
thus, total odd nos = (201-101) =100
correct me if Iam wrong
Hi Kabir:
It's ok:100 numbers.
Using another formula:
(299-101)/2 + 1 = 100
We got it right.
Silvia
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by grockit_jake » Mon Dec 07, 2009 1:48 pm
Personally, I find formulas to be an easy trap for test takers. There are a lot floating out there, and while you need the essential ones, many times you end up memorize a lot of stuff you may not need and it gets muddled in the anxiety of the real test.

By no means am I discouraging the use of formulas, especially in a case like this. If you are comfortable with applying a standard operating procedure to a particular type of problem, that will certainly save you time and effort. In general, however, be care on the test that you aren't

a. using the wrong formula by accident
b. omit or add a tiny piece
c. apply formulas to situations that are not very complicated and do not require it.

Instead, for both easy and difficult questions, supplement or substitute the use of a formula with a quick pattern discovery.

For either of these sum questions, you'll need to know the average of the first + last, and the total number pairs. From there, you can generally reason the answer without a set formula.
Jake Becker
Academic Director
Grockit Test Prep
https://www.grockit.com
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by valleeny » Tue Dec 15, 2009 11:20 pm
I spent hours and managed to convince myself of all concept spoken here and I am replying here to drill the concept so that such questions will never be an issue again.

1) For any given set of numbers, first we need to redefine the range that the question is asking for.
Eg. Sum of all even numbers from 99 to 301.
The range of numbers we need to work with should be redefined as 100 to 300.

2) Remember this. For a set with evenly spaced numbers (ie numbers that differ from the ones before and after by the same number, visualize 1,2,3,4 is evenly spaced with 1 the difference between the numbers, and 2,4,6,8 evenly spaced with 2 the difference between the numbers), the average is simply
Average = (Last number + First number)/2
This average applies as long as numbers are evenly spaced in the range.

3) No of terms (doesn't matter even or odd) = ((Last number - First number)/2) + 1
(Just remember this formula. You can try to reason it and convince yourself once and for all. I took very long to convince myself)

4) Sum = Average x No. of terms


Using our example question with steps below
Question : Sum of all even numbers from 99 to 301
1) Redefine range : 100 to 300
2) Numbers are evenly spaced. Average = (100+300)/2 = 200
3) No. of even terms in redefined range = ((300-100)/2)+1 = 101
4) Sum of all even numbers = 200 x 101 = 20200

Question : Sum of all even numbers from 1 to 199
1) Redefine range : 2 to 198
2) Average = (198+2)/2 = 100
3) No. of terms = ((198-2)/2)+1 = 99
4) Sum = 99 x 100 = 9900

Question : Sum of all odd numbers from 100 to 300
1) Redefine range : 101 to 299
2) Average = (299+101)/2 = 200
3) No. of terms = ((299-101)/2)+1 = 100
4) Sum = 100 x 200 = 20000

With this steps I think I wouldn't miss another such question. Let me know if I'm right.
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by benjiboo » Wed Dec 16, 2009 10:27 am
Properties of Evenly Spaced Sets
In an evenly spaced set (aka an arithmetic sequence):

1. The mean (average) and the median are equal to each other
2. The mean (average) and the median are equal to the average of the FIRST and LAST terms
3. The SUM of the elements in the set is the mean (average) times the number of terms in the set

How to find the number of terms in a consecutive set
How many terms are there from 5-20? The answer is not 15, but rather 16. Here are the rules for finding the number of terms in a set.

1. Number of terms in a consecutive set = (Last Term - First Term + 1)
2. Number of even (or odd terms) in a consecutive set = (Last Term - First Term)/2 + 1
3. Number of consecutive multiples in a consecutive set = (Last Term - First Term)/increment + 1 where the increment is the multiple.

IMPORTANT: Notice how I used the word "term" and not number. This is important because sometimes you don't always just put the first and last number you are given. For example, If you are asked to find the number of even integers between 1 and 30, you don't use the "first number" in the set. The first number is "1", which is odd, and we are only speaking about even numbers. Therefore, the first term is "2", not "1", even though the set or question might have stated "from 1-30". Same goes with the last term.

Sample Questions and How To Answer

Find the sum of integers from 20 to 100
1. Find the average of the set (First Term + Last Term)/2 (20+100)/2=60
2. Find the number of terms (Last Term - First Term) + 1 (100-20)+1=81
3. Multiple the average by the number of terms to get answer 60*81=4860

Find the sum of the EVEN integers from 20 to 100
1. Find the average of the set (First Term + Last Term)/2 (20+100)/2=60
2. Find the number of terms (Last Term - First Term)/2 + 1 (100-20)/2 +1=41
3. Multiple the average by the number of terms to get answer 60*41=2460

Find the sum of the ODD integers from 20 to 100
1. Find the average of the set (First Term + Last Term)/2 (21+99)/2=60
2. Find the number of terms (Last Term - First Term)/2 + 1 (99-21)/2 +1= 40
3. Multiple the average by the number of terms to get answer 60*40=2400

"¢ Notice that the average of an odd number of consecutive integers will be an integer whereas if there are an even amount the average will not be an integer but will be between two integers (and always end in .5)
"¢ Also note that the SUM of a set of consecutive integers with an ODD number of terms will be a multiple of the number of terms.
"¢ The SUM of a set of consecutive integers with an EVEN number of terms will never be a multiple of the number of terms. Check this out by looking at the examples above.
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